Find $lim _{xto infty }x^{frac{1}{x}}$ without L'Hôpital's Rule












-1












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$$lim _{xto infty }x^{frac{1}{x}}$$ How can I solve this? I want the most simple way to do it. Is there a nice log rule I can use here? I'm definitely not at the stage of using the limit chain rule as suggested on SymboLab.



I couldn't get anywhere with it using the methods I know ($0/0, a/0, a/infty,$ basic logs).



Edit: This is an extremely snaky move from me to add the limits-without-lhopital tag now. Is it too late, or do I need to add a separate question?










share|cite|improve this question











$endgroup$

















    -1












    $begingroup$


    $$lim _{xto infty }x^{frac{1}{x}}$$ How can I solve this? I want the most simple way to do it. Is there a nice log rule I can use here? I'm definitely not at the stage of using the limit chain rule as suggested on SymboLab.



    I couldn't get anywhere with it using the methods I know ($0/0, a/0, a/infty,$ basic logs).



    Edit: This is an extremely snaky move from me to add the limits-without-lhopital tag now. Is it too late, or do I need to add a separate question?










    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1


      1



      $begingroup$


      $$lim _{xto infty }x^{frac{1}{x}}$$ How can I solve this? I want the most simple way to do it. Is there a nice log rule I can use here? I'm definitely not at the stage of using the limit chain rule as suggested on SymboLab.



      I couldn't get anywhere with it using the methods I know ($0/0, a/0, a/infty,$ basic logs).



      Edit: This is an extremely snaky move from me to add the limits-without-lhopital tag now. Is it too late, or do I need to add a separate question?










      share|cite|improve this question











      $endgroup$




      $$lim _{xto infty }x^{frac{1}{x}}$$ How can I solve this? I want the most simple way to do it. Is there a nice log rule I can use here? I'm definitely not at the stage of using the limit chain rule as suggested on SymboLab.



      I couldn't get anywhere with it using the methods I know ($0/0, a/0, a/infty,$ basic logs).



      Edit: This is an extremely snaky move from me to add the limits-without-lhopital tag now. Is it too late, or do I need to add a separate question?







      limits limits-without-lhopital






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jul 27 '16 at 19:16







      Max Li

















      asked Jul 27 '16 at 19:06









      Max LiMax Li

      860619




      860619






















          6 Answers
          6






          active

          oldest

          votes


















          5












          $begingroup$

          If you know that for any $;epsilon>0;$ we have that $;log x<x^epsilon;$ for any $;0<xinBbb R;$ big enough , you then don't need l'Hospital since we can write



          $$frac{log x}x<frac{x^epsilon}x=frac1{x^{1-epsilon}}xrightarrow[xtoinfty]{}0;,;;text{for say};;0<epsilon<1$$



          Using that $;log x>log1=0;$ for $;x>1;$, use the squeeze theorem to deduce that



          $$lim_{xtoinfty}frac{log x}x=0$$



          and, as in the other answers, use now the algebraic equality



          $$x^{1/x}=e^{frac1xlog x}implieslim_{xtoinfty}x^{1/x}=e^{limlimits_{xtoinfty}frac1xlog x}=e^0=1$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            nice answer,without l'hospital rule ,+1
            $endgroup$
            – haqnatural
            Jul 27 '16 at 19:28










          • $begingroup$
            What is an elementary way to show the first claim (that $log x = O(x^epsilon)$)?
            $endgroup$
            – angryavian
            Jul 27 '16 at 19:48










          • $begingroup$
            @angryavian I'm not sure. I think I can prove it by means of differentials but I'm not sure if this makes the cut of "elementary".
            $endgroup$
            – DonAntonio
            Jul 27 '16 at 19:56










          • $begingroup$
            @DonAntonio I guess I meant (ideally) without calculus, or at at least something at the same level as the rest of your answer. Is there something along the lines of this answer?
            $endgroup$
            – angryavian
            Jul 27 '16 at 20:22



















          7












          $begingroup$

          firstly,write it as $x^{ frac { 1 }{ x } }={ e }^{ frac { 1 }{ x } ln { x } }$
          now we have such limit $$lim _{ xto infty } e^{ frac { ln { x } }{ x } }=lim _{ xrightarrow infty }{ { e }^{ frac { fleft( x right) rightarrow infty }{ gleft( x right) rightarrow infty } } } ,quad wherequad fleft( x right) =ln { x } ,gleft( x right) =x $$
          so we can apply L'hospital's rule




          $$lim _{ xto infty } x^{ frac { 1 }{ x } }=lim _{ xto infty } e^{ frac { 1 }{ x } ln { x } }=lim _{ xto infty } e^{ frac { 1 }{ x } }=1$$







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you do this step-by-step providing sources or explanations for the properties you have used?
            $endgroup$
            – Max Li
            Jul 27 '16 at 19:10






          • 1




            $begingroup$
            @MaxLi The first two are algaebraically equal. The second and third involve L'Hopital's Rule. And the last is evaluating the limit.
            $endgroup$
            – Simply Beautiful Art
            Jul 27 '16 at 19:12










          • $begingroup$
            Is there a way without L'Hôpital's Rule?
            $endgroup$
            – Max Li
            Jul 27 '16 at 19:17










          • $begingroup$
            yes,you can prove that $x>ln { x } ,x>1$ so that x "runs" to infinity more quickly than $lnx$
            $endgroup$
            – haqnatural
            Jul 27 '16 at 19:20






          • 1




            $begingroup$
            @Bernard Indeed so...but almost .
            $endgroup$
            – DonAntonio
            Jul 27 '16 at 19:25



















          3












          $begingroup$

          $x^{1/x}=e^{ln(x)/x}$ so the limit is $limlimits_{xrightarrow +infty}e^{ln(x)/x}$.

          You know that $limlimits_{+infty}ln(x)/x=0$ this implies that $limlimits_{xrightarrow +infty}e^{ln(x)/x}=1.$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you do this step-by-step providing sources or explanations for the properties you have used?
            $endgroup$
            – Max Li
            Jul 27 '16 at 19:10










          • $begingroup$
            Shouldn't it be $ln(x)$ not $log(x)$?
            $endgroup$
            – Max Li
            Jul 27 '16 at 19:18










          • $begingroup$
            @MaxLi $log = ln$ for most purposes. Also, passing the limit to the exponent is justified by continuity of $exp$.
            $endgroup$
            – Zain Patel
            Jul 27 '16 at 19:19










          • $begingroup$
            @MaxLi Under the right context (computer science) $log$ can imply $log_2$. Usually, $log=log_{10}$, and if we are doing limits/calculus, $log=log_e=ln$.
            $endgroup$
            – Simply Beautiful Art
            Jul 27 '16 at 19:32



















          1












          $begingroup$

          If you consider this
          for integer $n$,
          you want to show that
          $lim_{n to infty} n^{1/n}
          = 1
          $.



          Here's a bit of magic:



          By Bernoulli's inequality,
          $(1+frac1{sqrt{n}})^n
          ge 1+nfrac1{sqrt{n}}
          ge 1+sqrt{n}
          gt sqrt{n}
          = n^{1/2}
          $.



          Raising both sides
          to the $2/n$ power,
          $((1+frac1{sqrt{n}})^n)^{2/n}
          > (n^{1/2})^{2/n}
          $
          or
          $(1+frac1{sqrt{n}})^2
          > n^{1/n}
          $
          or
          $n^{1/n}
          < (1+frac1{sqrt{n}})^2
          =1+2frac1{sqrt{n}}+frac1{n}
          <1+3frac1{sqrt{n}}
          $.



          Since
          $n^{1/n} > 1$
          and
          $lim_{n to infty}frac1{sqrt{n}}
          =0
          $,
          $lim_{n to infty} n^{1/n}
          =
          1
          $.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            As to avoiding the use of L'Hospital's rule:



            $$ln(x)=O(sqrt x)$$



            $$implieslim_{xtoinfty}frac{ln(x)}x=lim_{xtoinfty}frac{sqrt x}x$$



            $$=lim_{xtoinfty}frac1{sqrt x}=0$$



            And so we have $lim x^{1/x}=e^{ln(x)/x}=e^0=1$






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              Let $x=1+frac {1}{y}$ Where $y rightarrow 0$



              Then,



              $lim _{xto infty }x^{frac{1}{x}} =lim _{yto 0 }{(1+frac {1}{y})}^{frac {y}{1+y}}
              = lim _{yto 0 }{(1+frac {1}{y})}^{yfrac {1}{1+y}}$



              $=lim _{y to 0}{1^{frac {1}{1+y}}}=1$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                The step from the third line to the last one is, to say the least, very questionable: how could you justify to take the limit of only a part of the expression in the right extreme of the line one before the last, but to leave other part still with $;y;$ ?
                $endgroup$
                – DonAntonio
                Jul 27 '16 at 19:58










              • $begingroup$
                Yes, the step before the last step should be $={1^{frac {1}{1}}}$
                $endgroup$
                – Pentapolis
                Jul 27 '16 at 20:50












              • $begingroup$
                If $yto 0$, then $1/ytoinfty$. Then in the end you have: $1+infty$, which is evidently not $1$.
                $endgroup$
                – Physicist137
                Aug 9 '16 at 18:39












              • $begingroup$
                $lim _{yto 0 }{(1+frac {1}{y})}^y=1$ @Physicist137 this limit is well known which is equal to 1.
                $endgroup$
                – Pentapolis
                Aug 9 '16 at 22:09










              • $begingroup$
                Oh! Obviously. You are right. My mistake. Sorry... =). And +1. =).
                $endgroup$
                – Physicist137
                Aug 10 '16 at 1:05














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              6 Answers
              6






              active

              oldest

              votes








              6 Answers
              6






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              5












              $begingroup$

              If you know that for any $;epsilon>0;$ we have that $;log x<x^epsilon;$ for any $;0<xinBbb R;$ big enough , you then don't need l'Hospital since we can write



              $$frac{log x}x<frac{x^epsilon}x=frac1{x^{1-epsilon}}xrightarrow[xtoinfty]{}0;,;;text{for say};;0<epsilon<1$$



              Using that $;log x>log1=0;$ for $;x>1;$, use the squeeze theorem to deduce that



              $$lim_{xtoinfty}frac{log x}x=0$$



              and, as in the other answers, use now the algebraic equality



              $$x^{1/x}=e^{frac1xlog x}implieslim_{xtoinfty}x^{1/x}=e^{limlimits_{xtoinfty}frac1xlog x}=e^0=1$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                nice answer,without l'hospital rule ,+1
                $endgroup$
                – haqnatural
                Jul 27 '16 at 19:28










              • $begingroup$
                What is an elementary way to show the first claim (that $log x = O(x^epsilon)$)?
                $endgroup$
                – angryavian
                Jul 27 '16 at 19:48










              • $begingroup$
                @angryavian I'm not sure. I think I can prove it by means of differentials but I'm not sure if this makes the cut of "elementary".
                $endgroup$
                – DonAntonio
                Jul 27 '16 at 19:56










              • $begingroup$
                @DonAntonio I guess I meant (ideally) without calculus, or at at least something at the same level as the rest of your answer. Is there something along the lines of this answer?
                $endgroup$
                – angryavian
                Jul 27 '16 at 20:22
















              5












              $begingroup$

              If you know that for any $;epsilon>0;$ we have that $;log x<x^epsilon;$ for any $;0<xinBbb R;$ big enough , you then don't need l'Hospital since we can write



              $$frac{log x}x<frac{x^epsilon}x=frac1{x^{1-epsilon}}xrightarrow[xtoinfty]{}0;,;;text{for say};;0<epsilon<1$$



              Using that $;log x>log1=0;$ for $;x>1;$, use the squeeze theorem to deduce that



              $$lim_{xtoinfty}frac{log x}x=0$$



              and, as in the other answers, use now the algebraic equality



              $$x^{1/x}=e^{frac1xlog x}implieslim_{xtoinfty}x^{1/x}=e^{limlimits_{xtoinfty}frac1xlog x}=e^0=1$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                nice answer,without l'hospital rule ,+1
                $endgroup$
                – haqnatural
                Jul 27 '16 at 19:28










              • $begingroup$
                What is an elementary way to show the first claim (that $log x = O(x^epsilon)$)?
                $endgroup$
                – angryavian
                Jul 27 '16 at 19:48










              • $begingroup$
                @angryavian I'm not sure. I think I can prove it by means of differentials but I'm not sure if this makes the cut of "elementary".
                $endgroup$
                – DonAntonio
                Jul 27 '16 at 19:56










              • $begingroup$
                @DonAntonio I guess I meant (ideally) without calculus, or at at least something at the same level as the rest of your answer. Is there something along the lines of this answer?
                $endgroup$
                – angryavian
                Jul 27 '16 at 20:22














              5












              5








              5





              $begingroup$

              If you know that for any $;epsilon>0;$ we have that $;log x<x^epsilon;$ for any $;0<xinBbb R;$ big enough , you then don't need l'Hospital since we can write



              $$frac{log x}x<frac{x^epsilon}x=frac1{x^{1-epsilon}}xrightarrow[xtoinfty]{}0;,;;text{for say};;0<epsilon<1$$



              Using that $;log x>log1=0;$ for $;x>1;$, use the squeeze theorem to deduce that



              $$lim_{xtoinfty}frac{log x}x=0$$



              and, as in the other answers, use now the algebraic equality



              $$x^{1/x}=e^{frac1xlog x}implieslim_{xtoinfty}x^{1/x}=e^{limlimits_{xtoinfty}frac1xlog x}=e^0=1$$






              share|cite|improve this answer











              $endgroup$



              If you know that for any $;epsilon>0;$ we have that $;log x<x^epsilon;$ for any $;0<xinBbb R;$ big enough , you then don't need l'Hospital since we can write



              $$frac{log x}x<frac{x^epsilon}x=frac1{x^{1-epsilon}}xrightarrow[xtoinfty]{}0;,;;text{for say};;0<epsilon<1$$



              Using that $;log x>log1=0;$ for $;x>1;$, use the squeeze theorem to deduce that



              $$lim_{xtoinfty}frac{log x}x=0$$



              and, as in the other answers, use now the algebraic equality



              $$x^{1/x}=e^{frac1xlog x}implieslim_{xtoinfty}x^{1/x}=e^{limlimits_{xtoinfty}frac1xlog x}=e^0=1$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jul 27 '16 at 19:29

























              answered Jul 27 '16 at 19:24









              DonAntonioDonAntonio

              180k1494233




              180k1494233












              • $begingroup$
                nice answer,without l'hospital rule ,+1
                $endgroup$
                – haqnatural
                Jul 27 '16 at 19:28










              • $begingroup$
                What is an elementary way to show the first claim (that $log x = O(x^epsilon)$)?
                $endgroup$
                – angryavian
                Jul 27 '16 at 19:48










              • $begingroup$
                @angryavian I'm not sure. I think I can prove it by means of differentials but I'm not sure if this makes the cut of "elementary".
                $endgroup$
                – DonAntonio
                Jul 27 '16 at 19:56










              • $begingroup$
                @DonAntonio I guess I meant (ideally) without calculus, or at at least something at the same level as the rest of your answer. Is there something along the lines of this answer?
                $endgroup$
                – angryavian
                Jul 27 '16 at 20:22


















              • $begingroup$
                nice answer,without l'hospital rule ,+1
                $endgroup$
                – haqnatural
                Jul 27 '16 at 19:28










              • $begingroup$
                What is an elementary way to show the first claim (that $log x = O(x^epsilon)$)?
                $endgroup$
                – angryavian
                Jul 27 '16 at 19:48










              • $begingroup$
                @angryavian I'm not sure. I think I can prove it by means of differentials but I'm not sure if this makes the cut of "elementary".
                $endgroup$
                – DonAntonio
                Jul 27 '16 at 19:56










              • $begingroup$
                @DonAntonio I guess I meant (ideally) without calculus, or at at least something at the same level as the rest of your answer. Is there something along the lines of this answer?
                $endgroup$
                – angryavian
                Jul 27 '16 at 20:22
















              $begingroup$
              nice answer,without l'hospital rule ,+1
              $endgroup$
              – haqnatural
              Jul 27 '16 at 19:28




              $begingroup$
              nice answer,without l'hospital rule ,+1
              $endgroup$
              – haqnatural
              Jul 27 '16 at 19:28












              $begingroup$
              What is an elementary way to show the first claim (that $log x = O(x^epsilon)$)?
              $endgroup$
              – angryavian
              Jul 27 '16 at 19:48




              $begingroup$
              What is an elementary way to show the first claim (that $log x = O(x^epsilon)$)?
              $endgroup$
              – angryavian
              Jul 27 '16 at 19:48












              $begingroup$
              @angryavian I'm not sure. I think I can prove it by means of differentials but I'm not sure if this makes the cut of "elementary".
              $endgroup$
              – DonAntonio
              Jul 27 '16 at 19:56




              $begingroup$
              @angryavian I'm not sure. I think I can prove it by means of differentials but I'm not sure if this makes the cut of "elementary".
              $endgroup$
              – DonAntonio
              Jul 27 '16 at 19:56












              $begingroup$
              @DonAntonio I guess I meant (ideally) without calculus, or at at least something at the same level as the rest of your answer. Is there something along the lines of this answer?
              $endgroup$
              – angryavian
              Jul 27 '16 at 20:22




              $begingroup$
              @DonAntonio I guess I meant (ideally) without calculus, or at at least something at the same level as the rest of your answer. Is there something along the lines of this answer?
              $endgroup$
              – angryavian
              Jul 27 '16 at 20:22











              7












              $begingroup$

              firstly,write it as $x^{ frac { 1 }{ x } }={ e }^{ frac { 1 }{ x } ln { x } }$
              now we have such limit $$lim _{ xto infty } e^{ frac { ln { x } }{ x } }=lim _{ xrightarrow infty }{ { e }^{ frac { fleft( x right) rightarrow infty }{ gleft( x right) rightarrow infty } } } ,quad wherequad fleft( x right) =ln { x } ,gleft( x right) =x $$
              so we can apply L'hospital's rule




              $$lim _{ xto infty } x^{ frac { 1 }{ x } }=lim _{ xto infty } e^{ frac { 1 }{ x } ln { x } }=lim _{ xto infty } e^{ frac { 1 }{ x } }=1$$







              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Can you do this step-by-step providing sources or explanations for the properties you have used?
                $endgroup$
                – Max Li
                Jul 27 '16 at 19:10






              • 1




                $begingroup$
                @MaxLi The first two are algaebraically equal. The second and third involve L'Hopital's Rule. And the last is evaluating the limit.
                $endgroup$
                – Simply Beautiful Art
                Jul 27 '16 at 19:12










              • $begingroup$
                Is there a way without L'Hôpital's Rule?
                $endgroup$
                – Max Li
                Jul 27 '16 at 19:17










              • $begingroup$
                yes,you can prove that $x>ln { x } ,x>1$ so that x "runs" to infinity more quickly than $lnx$
                $endgroup$
                – haqnatural
                Jul 27 '16 at 19:20






              • 1




                $begingroup$
                @Bernard Indeed so...but almost .
                $endgroup$
                – DonAntonio
                Jul 27 '16 at 19:25
















              7












              $begingroup$

              firstly,write it as $x^{ frac { 1 }{ x } }={ e }^{ frac { 1 }{ x } ln { x } }$
              now we have such limit $$lim _{ xto infty } e^{ frac { ln { x } }{ x } }=lim _{ xrightarrow infty }{ { e }^{ frac { fleft( x right) rightarrow infty }{ gleft( x right) rightarrow infty } } } ,quad wherequad fleft( x right) =ln { x } ,gleft( x right) =x $$
              so we can apply L'hospital's rule




              $$lim _{ xto infty } x^{ frac { 1 }{ x } }=lim _{ xto infty } e^{ frac { 1 }{ x } ln { x } }=lim _{ xto infty } e^{ frac { 1 }{ x } }=1$$







              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Can you do this step-by-step providing sources or explanations for the properties you have used?
                $endgroup$
                – Max Li
                Jul 27 '16 at 19:10






              • 1




                $begingroup$
                @MaxLi The first two are algaebraically equal. The second and third involve L'Hopital's Rule. And the last is evaluating the limit.
                $endgroup$
                – Simply Beautiful Art
                Jul 27 '16 at 19:12










              • $begingroup$
                Is there a way without L'Hôpital's Rule?
                $endgroup$
                – Max Li
                Jul 27 '16 at 19:17










              • $begingroup$
                yes,you can prove that $x>ln { x } ,x>1$ so that x "runs" to infinity more quickly than $lnx$
                $endgroup$
                – haqnatural
                Jul 27 '16 at 19:20






              • 1




                $begingroup$
                @Bernard Indeed so...but almost .
                $endgroup$
                – DonAntonio
                Jul 27 '16 at 19:25














              7












              7








              7





              $begingroup$

              firstly,write it as $x^{ frac { 1 }{ x } }={ e }^{ frac { 1 }{ x } ln { x } }$
              now we have such limit $$lim _{ xto infty } e^{ frac { ln { x } }{ x } }=lim _{ xrightarrow infty }{ { e }^{ frac { fleft( x right) rightarrow infty }{ gleft( x right) rightarrow infty } } } ,quad wherequad fleft( x right) =ln { x } ,gleft( x right) =x $$
              so we can apply L'hospital's rule




              $$lim _{ xto infty } x^{ frac { 1 }{ x } }=lim _{ xto infty } e^{ frac { 1 }{ x } ln { x } }=lim _{ xto infty } e^{ frac { 1 }{ x } }=1$$







              share|cite|improve this answer











              $endgroup$



              firstly,write it as $x^{ frac { 1 }{ x } }={ e }^{ frac { 1 }{ x } ln { x } }$
              now we have such limit $$lim _{ xto infty } e^{ frac { ln { x } }{ x } }=lim _{ xrightarrow infty }{ { e }^{ frac { fleft( x right) rightarrow infty }{ gleft( x right) rightarrow infty } } } ,quad wherequad fleft( x right) =ln { x } ,gleft( x right) =x $$
              so we can apply L'hospital's rule




              $$lim _{ xto infty } x^{ frac { 1 }{ x } }=lim _{ xto infty } e^{ frac { 1 }{ x } ln { x } }=lim _{ xto infty } e^{ frac { 1 }{ x } }=1$$








              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jul 27 '16 at 19:14

























              answered Jul 27 '16 at 19:08









              haqnaturalhaqnatural

              20.8k72457




              20.8k72457












              • $begingroup$
                Can you do this step-by-step providing sources or explanations for the properties you have used?
                $endgroup$
                – Max Li
                Jul 27 '16 at 19:10






              • 1




                $begingroup$
                @MaxLi The first two are algaebraically equal. The second and third involve L'Hopital's Rule. And the last is evaluating the limit.
                $endgroup$
                – Simply Beautiful Art
                Jul 27 '16 at 19:12










              • $begingroup$
                Is there a way without L'Hôpital's Rule?
                $endgroup$
                – Max Li
                Jul 27 '16 at 19:17










              • $begingroup$
                yes,you can prove that $x>ln { x } ,x>1$ so that x "runs" to infinity more quickly than $lnx$
                $endgroup$
                – haqnatural
                Jul 27 '16 at 19:20






              • 1




                $begingroup$
                @Bernard Indeed so...but almost .
                $endgroup$
                – DonAntonio
                Jul 27 '16 at 19:25


















              • $begingroup$
                Can you do this step-by-step providing sources or explanations for the properties you have used?
                $endgroup$
                – Max Li
                Jul 27 '16 at 19:10






              • 1




                $begingroup$
                @MaxLi The first two are algaebraically equal. The second and third involve L'Hopital's Rule. And the last is evaluating the limit.
                $endgroup$
                – Simply Beautiful Art
                Jul 27 '16 at 19:12










              • $begingroup$
                Is there a way without L'Hôpital's Rule?
                $endgroup$
                – Max Li
                Jul 27 '16 at 19:17










              • $begingroup$
                yes,you can prove that $x>ln { x } ,x>1$ so that x "runs" to infinity more quickly than $lnx$
                $endgroup$
                – haqnatural
                Jul 27 '16 at 19:20






              • 1




                $begingroup$
                @Bernard Indeed so...but almost .
                $endgroup$
                – DonAntonio
                Jul 27 '16 at 19:25
















              $begingroup$
              Can you do this step-by-step providing sources or explanations for the properties you have used?
              $endgroup$
              – Max Li
              Jul 27 '16 at 19:10




              $begingroup$
              Can you do this step-by-step providing sources or explanations for the properties you have used?
              $endgroup$
              – Max Li
              Jul 27 '16 at 19:10




              1




              1




              $begingroup$
              @MaxLi The first two are algaebraically equal. The second and third involve L'Hopital's Rule. And the last is evaluating the limit.
              $endgroup$
              – Simply Beautiful Art
              Jul 27 '16 at 19:12




              $begingroup$
              @MaxLi The first two are algaebraically equal. The second and third involve L'Hopital's Rule. And the last is evaluating the limit.
              $endgroup$
              – Simply Beautiful Art
              Jul 27 '16 at 19:12












              $begingroup$
              Is there a way without L'Hôpital's Rule?
              $endgroup$
              – Max Li
              Jul 27 '16 at 19:17




              $begingroup$
              Is there a way without L'Hôpital's Rule?
              $endgroup$
              – Max Li
              Jul 27 '16 at 19:17












              $begingroup$
              yes,you can prove that $x>ln { x } ,x>1$ so that x "runs" to infinity more quickly than $lnx$
              $endgroup$
              – haqnatural
              Jul 27 '16 at 19:20




              $begingroup$
              yes,you can prove that $x>ln { x } ,x>1$ so that x "runs" to infinity more quickly than $lnx$
              $endgroup$
              – haqnatural
              Jul 27 '16 at 19:20




              1




              1




              $begingroup$
              @Bernard Indeed so...but almost .
              $endgroup$
              – DonAntonio
              Jul 27 '16 at 19:25




              $begingroup$
              @Bernard Indeed so...but almost .
              $endgroup$
              – DonAntonio
              Jul 27 '16 at 19:25











              3












              $begingroup$

              $x^{1/x}=e^{ln(x)/x}$ so the limit is $limlimits_{xrightarrow +infty}e^{ln(x)/x}$.

              You know that $limlimits_{+infty}ln(x)/x=0$ this implies that $limlimits_{xrightarrow +infty}e^{ln(x)/x}=1.$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Can you do this step-by-step providing sources or explanations for the properties you have used?
                $endgroup$
                – Max Li
                Jul 27 '16 at 19:10










              • $begingroup$
                Shouldn't it be $ln(x)$ not $log(x)$?
                $endgroup$
                – Max Li
                Jul 27 '16 at 19:18










              • $begingroup$
                @MaxLi $log = ln$ for most purposes. Also, passing the limit to the exponent is justified by continuity of $exp$.
                $endgroup$
                – Zain Patel
                Jul 27 '16 at 19:19










              • $begingroup$
                @MaxLi Under the right context (computer science) $log$ can imply $log_2$. Usually, $log=log_{10}$, and if we are doing limits/calculus, $log=log_e=ln$.
                $endgroup$
                – Simply Beautiful Art
                Jul 27 '16 at 19:32
















              3












              $begingroup$

              $x^{1/x}=e^{ln(x)/x}$ so the limit is $limlimits_{xrightarrow +infty}e^{ln(x)/x}$.

              You know that $limlimits_{+infty}ln(x)/x=0$ this implies that $limlimits_{xrightarrow +infty}e^{ln(x)/x}=1.$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Can you do this step-by-step providing sources or explanations for the properties you have used?
                $endgroup$
                – Max Li
                Jul 27 '16 at 19:10










              • $begingroup$
                Shouldn't it be $ln(x)$ not $log(x)$?
                $endgroup$
                – Max Li
                Jul 27 '16 at 19:18










              • $begingroup$
                @MaxLi $log = ln$ for most purposes. Also, passing the limit to the exponent is justified by continuity of $exp$.
                $endgroup$
                – Zain Patel
                Jul 27 '16 at 19:19










              • $begingroup$
                @MaxLi Under the right context (computer science) $log$ can imply $log_2$. Usually, $log=log_{10}$, and if we are doing limits/calculus, $log=log_e=ln$.
                $endgroup$
                – Simply Beautiful Art
                Jul 27 '16 at 19:32














              3












              3








              3





              $begingroup$

              $x^{1/x}=e^{ln(x)/x}$ so the limit is $limlimits_{xrightarrow +infty}e^{ln(x)/x}$.

              You know that $limlimits_{+infty}ln(x)/x=0$ this implies that $limlimits_{xrightarrow +infty}e^{ln(x)/x}=1.$






              share|cite|improve this answer











              $endgroup$



              $x^{1/x}=e^{ln(x)/x}$ so the limit is $limlimits_{xrightarrow +infty}e^{ln(x)/x}$.

              You know that $limlimits_{+infty}ln(x)/x=0$ this implies that $limlimits_{xrightarrow +infty}e^{ln(x)/x}=1.$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 16 at 16:42









              user376343

              3,9584829




              3,9584829










              answered Jul 27 '16 at 19:08









              Tsemo AristideTsemo Aristide

              60.2k11446




              60.2k11446












              • $begingroup$
                Can you do this step-by-step providing sources or explanations for the properties you have used?
                $endgroup$
                – Max Li
                Jul 27 '16 at 19:10










              • $begingroup$
                Shouldn't it be $ln(x)$ not $log(x)$?
                $endgroup$
                – Max Li
                Jul 27 '16 at 19:18










              • $begingroup$
                @MaxLi $log = ln$ for most purposes. Also, passing the limit to the exponent is justified by continuity of $exp$.
                $endgroup$
                – Zain Patel
                Jul 27 '16 at 19:19










              • $begingroup$
                @MaxLi Under the right context (computer science) $log$ can imply $log_2$. Usually, $log=log_{10}$, and if we are doing limits/calculus, $log=log_e=ln$.
                $endgroup$
                – Simply Beautiful Art
                Jul 27 '16 at 19:32


















              • $begingroup$
                Can you do this step-by-step providing sources or explanations for the properties you have used?
                $endgroup$
                – Max Li
                Jul 27 '16 at 19:10










              • $begingroup$
                Shouldn't it be $ln(x)$ not $log(x)$?
                $endgroup$
                – Max Li
                Jul 27 '16 at 19:18










              • $begingroup$
                @MaxLi $log = ln$ for most purposes. Also, passing the limit to the exponent is justified by continuity of $exp$.
                $endgroup$
                – Zain Patel
                Jul 27 '16 at 19:19










              • $begingroup$
                @MaxLi Under the right context (computer science) $log$ can imply $log_2$. Usually, $log=log_{10}$, and if we are doing limits/calculus, $log=log_e=ln$.
                $endgroup$
                – Simply Beautiful Art
                Jul 27 '16 at 19:32
















              $begingroup$
              Can you do this step-by-step providing sources or explanations for the properties you have used?
              $endgroup$
              – Max Li
              Jul 27 '16 at 19:10




              $begingroup$
              Can you do this step-by-step providing sources or explanations for the properties you have used?
              $endgroup$
              – Max Li
              Jul 27 '16 at 19:10












              $begingroup$
              Shouldn't it be $ln(x)$ not $log(x)$?
              $endgroup$
              – Max Li
              Jul 27 '16 at 19:18




              $begingroup$
              Shouldn't it be $ln(x)$ not $log(x)$?
              $endgroup$
              – Max Li
              Jul 27 '16 at 19:18












              $begingroup$
              @MaxLi $log = ln$ for most purposes. Also, passing the limit to the exponent is justified by continuity of $exp$.
              $endgroup$
              – Zain Patel
              Jul 27 '16 at 19:19




              $begingroup$
              @MaxLi $log = ln$ for most purposes. Also, passing the limit to the exponent is justified by continuity of $exp$.
              $endgroup$
              – Zain Patel
              Jul 27 '16 at 19:19












              $begingroup$
              @MaxLi Under the right context (computer science) $log$ can imply $log_2$. Usually, $log=log_{10}$, and if we are doing limits/calculus, $log=log_e=ln$.
              $endgroup$
              – Simply Beautiful Art
              Jul 27 '16 at 19:32




              $begingroup$
              @MaxLi Under the right context (computer science) $log$ can imply $log_2$. Usually, $log=log_{10}$, and if we are doing limits/calculus, $log=log_e=ln$.
              $endgroup$
              – Simply Beautiful Art
              Jul 27 '16 at 19:32











              1












              $begingroup$

              If you consider this
              for integer $n$,
              you want to show that
              $lim_{n to infty} n^{1/n}
              = 1
              $.



              Here's a bit of magic:



              By Bernoulli's inequality,
              $(1+frac1{sqrt{n}})^n
              ge 1+nfrac1{sqrt{n}}
              ge 1+sqrt{n}
              gt sqrt{n}
              = n^{1/2}
              $.



              Raising both sides
              to the $2/n$ power,
              $((1+frac1{sqrt{n}})^n)^{2/n}
              > (n^{1/2})^{2/n}
              $
              or
              $(1+frac1{sqrt{n}})^2
              > n^{1/n}
              $
              or
              $n^{1/n}
              < (1+frac1{sqrt{n}})^2
              =1+2frac1{sqrt{n}}+frac1{n}
              <1+3frac1{sqrt{n}}
              $.



              Since
              $n^{1/n} > 1$
              and
              $lim_{n to infty}frac1{sqrt{n}}
              =0
              $,
              $lim_{n to infty} n^{1/n}
              =
              1
              $.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                If you consider this
                for integer $n$,
                you want to show that
                $lim_{n to infty} n^{1/n}
                = 1
                $.



                Here's a bit of magic:



                By Bernoulli's inequality,
                $(1+frac1{sqrt{n}})^n
                ge 1+nfrac1{sqrt{n}}
                ge 1+sqrt{n}
                gt sqrt{n}
                = n^{1/2}
                $.



                Raising both sides
                to the $2/n$ power,
                $((1+frac1{sqrt{n}})^n)^{2/n}
                > (n^{1/2})^{2/n}
                $
                or
                $(1+frac1{sqrt{n}})^2
                > n^{1/n}
                $
                or
                $n^{1/n}
                < (1+frac1{sqrt{n}})^2
                =1+2frac1{sqrt{n}}+frac1{n}
                <1+3frac1{sqrt{n}}
                $.



                Since
                $n^{1/n} > 1$
                and
                $lim_{n to infty}frac1{sqrt{n}}
                =0
                $,
                $lim_{n to infty} n^{1/n}
                =
                1
                $.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  If you consider this
                  for integer $n$,
                  you want to show that
                  $lim_{n to infty} n^{1/n}
                  = 1
                  $.



                  Here's a bit of magic:



                  By Bernoulli's inequality,
                  $(1+frac1{sqrt{n}})^n
                  ge 1+nfrac1{sqrt{n}}
                  ge 1+sqrt{n}
                  gt sqrt{n}
                  = n^{1/2}
                  $.



                  Raising both sides
                  to the $2/n$ power,
                  $((1+frac1{sqrt{n}})^n)^{2/n}
                  > (n^{1/2})^{2/n}
                  $
                  or
                  $(1+frac1{sqrt{n}})^2
                  > n^{1/n}
                  $
                  or
                  $n^{1/n}
                  < (1+frac1{sqrt{n}})^2
                  =1+2frac1{sqrt{n}}+frac1{n}
                  <1+3frac1{sqrt{n}}
                  $.



                  Since
                  $n^{1/n} > 1$
                  and
                  $lim_{n to infty}frac1{sqrt{n}}
                  =0
                  $,
                  $lim_{n to infty} n^{1/n}
                  =
                  1
                  $.






                  share|cite|improve this answer









                  $endgroup$



                  If you consider this
                  for integer $n$,
                  you want to show that
                  $lim_{n to infty} n^{1/n}
                  = 1
                  $.



                  Here's a bit of magic:



                  By Bernoulli's inequality,
                  $(1+frac1{sqrt{n}})^n
                  ge 1+nfrac1{sqrt{n}}
                  ge 1+sqrt{n}
                  gt sqrt{n}
                  = n^{1/2}
                  $.



                  Raising both sides
                  to the $2/n$ power,
                  $((1+frac1{sqrt{n}})^n)^{2/n}
                  > (n^{1/2})^{2/n}
                  $
                  or
                  $(1+frac1{sqrt{n}})^2
                  > n^{1/n}
                  $
                  or
                  $n^{1/n}
                  < (1+frac1{sqrt{n}})^2
                  =1+2frac1{sqrt{n}}+frac1{n}
                  <1+3frac1{sqrt{n}}
                  $.



                  Since
                  $n^{1/n} > 1$
                  and
                  $lim_{n to infty}frac1{sqrt{n}}
                  =0
                  $,
                  $lim_{n to infty} n^{1/n}
                  =
                  1
                  $.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jul 27 '16 at 20:32









                  marty cohenmarty cohen

                  75k549130




                  75k549130























                      0












                      $begingroup$

                      As to avoiding the use of L'Hospital's rule:



                      $$ln(x)=O(sqrt x)$$



                      $$implieslim_{xtoinfty}frac{ln(x)}x=lim_{xtoinfty}frac{sqrt x}x$$



                      $$=lim_{xtoinfty}frac1{sqrt x}=0$$



                      And so we have $lim x^{1/x}=e^{ln(x)/x}=e^0=1$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        As to avoiding the use of L'Hospital's rule:



                        $$ln(x)=O(sqrt x)$$



                        $$implieslim_{xtoinfty}frac{ln(x)}x=lim_{xtoinfty}frac{sqrt x}x$$



                        $$=lim_{xtoinfty}frac1{sqrt x}=0$$



                        And so we have $lim x^{1/x}=e^{ln(x)/x}=e^0=1$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          As to avoiding the use of L'Hospital's rule:



                          $$ln(x)=O(sqrt x)$$



                          $$implieslim_{xtoinfty}frac{ln(x)}x=lim_{xtoinfty}frac{sqrt x}x$$



                          $$=lim_{xtoinfty}frac1{sqrt x}=0$$



                          And so we have $lim x^{1/x}=e^{ln(x)/x}=e^0=1$






                          share|cite|improve this answer









                          $endgroup$



                          As to avoiding the use of L'Hospital's rule:



                          $$ln(x)=O(sqrt x)$$



                          $$implieslim_{xtoinfty}frac{ln(x)}x=lim_{xtoinfty}frac{sqrt x}x$$



                          $$=lim_{xtoinfty}frac1{sqrt x}=0$$



                          And so we have $lim x^{1/x}=e^{ln(x)/x}=e^0=1$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jul 27 '16 at 19:35









                          Simply Beautiful ArtSimply Beautiful Art

                          50.9k580186




                          50.9k580186























                              0












                              $begingroup$

                              Let $x=1+frac {1}{y}$ Where $y rightarrow 0$



                              Then,



                              $lim _{xto infty }x^{frac{1}{x}} =lim _{yto 0 }{(1+frac {1}{y})}^{frac {y}{1+y}}
                              = lim _{yto 0 }{(1+frac {1}{y})}^{yfrac {1}{1+y}}$



                              $=lim _{y to 0}{1^{frac {1}{1+y}}}=1$






                              share|cite|improve this answer









                              $endgroup$













                              • $begingroup$
                                The step from the third line to the last one is, to say the least, very questionable: how could you justify to take the limit of only a part of the expression in the right extreme of the line one before the last, but to leave other part still with $;y;$ ?
                                $endgroup$
                                – DonAntonio
                                Jul 27 '16 at 19:58










                              • $begingroup$
                                Yes, the step before the last step should be $={1^{frac {1}{1}}}$
                                $endgroup$
                                – Pentapolis
                                Jul 27 '16 at 20:50












                              • $begingroup$
                                If $yto 0$, then $1/ytoinfty$. Then in the end you have: $1+infty$, which is evidently not $1$.
                                $endgroup$
                                – Physicist137
                                Aug 9 '16 at 18:39












                              • $begingroup$
                                $lim _{yto 0 }{(1+frac {1}{y})}^y=1$ @Physicist137 this limit is well known which is equal to 1.
                                $endgroup$
                                – Pentapolis
                                Aug 9 '16 at 22:09










                              • $begingroup$
                                Oh! Obviously. You are right. My mistake. Sorry... =). And +1. =).
                                $endgroup$
                                – Physicist137
                                Aug 10 '16 at 1:05


















                              0












                              $begingroup$

                              Let $x=1+frac {1}{y}$ Where $y rightarrow 0$



                              Then,



                              $lim _{xto infty }x^{frac{1}{x}} =lim _{yto 0 }{(1+frac {1}{y})}^{frac {y}{1+y}}
                              = lim _{yto 0 }{(1+frac {1}{y})}^{yfrac {1}{1+y}}$



                              $=lim _{y to 0}{1^{frac {1}{1+y}}}=1$






                              share|cite|improve this answer









                              $endgroup$













                              • $begingroup$
                                The step from the third line to the last one is, to say the least, very questionable: how could you justify to take the limit of only a part of the expression in the right extreme of the line one before the last, but to leave other part still with $;y;$ ?
                                $endgroup$
                                – DonAntonio
                                Jul 27 '16 at 19:58










                              • $begingroup$
                                Yes, the step before the last step should be $={1^{frac {1}{1}}}$
                                $endgroup$
                                – Pentapolis
                                Jul 27 '16 at 20:50












                              • $begingroup$
                                If $yto 0$, then $1/ytoinfty$. Then in the end you have: $1+infty$, which is evidently not $1$.
                                $endgroup$
                                – Physicist137
                                Aug 9 '16 at 18:39












                              • $begingroup$
                                $lim _{yto 0 }{(1+frac {1}{y})}^y=1$ @Physicist137 this limit is well known which is equal to 1.
                                $endgroup$
                                – Pentapolis
                                Aug 9 '16 at 22:09










                              • $begingroup$
                                Oh! Obviously. You are right. My mistake. Sorry... =). And +1. =).
                                $endgroup$
                                – Physicist137
                                Aug 10 '16 at 1:05
















                              0












                              0








                              0





                              $begingroup$

                              Let $x=1+frac {1}{y}$ Where $y rightarrow 0$



                              Then,



                              $lim _{xto infty }x^{frac{1}{x}} =lim _{yto 0 }{(1+frac {1}{y})}^{frac {y}{1+y}}
                              = lim _{yto 0 }{(1+frac {1}{y})}^{yfrac {1}{1+y}}$



                              $=lim _{y to 0}{1^{frac {1}{1+y}}}=1$






                              share|cite|improve this answer









                              $endgroup$



                              Let $x=1+frac {1}{y}$ Where $y rightarrow 0$



                              Then,



                              $lim _{xto infty }x^{frac{1}{x}} =lim _{yto 0 }{(1+frac {1}{y})}^{frac {y}{1+y}}
                              = lim _{yto 0 }{(1+frac {1}{y})}^{yfrac {1}{1+y}}$



                              $=lim _{y to 0}{1^{frac {1}{1+y}}}=1$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jul 27 '16 at 19:49









                              PentapolisPentapolis

                              36419




                              36419












                              • $begingroup$
                                The step from the third line to the last one is, to say the least, very questionable: how could you justify to take the limit of only a part of the expression in the right extreme of the line one before the last, but to leave other part still with $;y;$ ?
                                $endgroup$
                                – DonAntonio
                                Jul 27 '16 at 19:58










                              • $begingroup$
                                Yes, the step before the last step should be $={1^{frac {1}{1}}}$
                                $endgroup$
                                – Pentapolis
                                Jul 27 '16 at 20:50












                              • $begingroup$
                                If $yto 0$, then $1/ytoinfty$. Then in the end you have: $1+infty$, which is evidently not $1$.
                                $endgroup$
                                – Physicist137
                                Aug 9 '16 at 18:39












                              • $begingroup$
                                $lim _{yto 0 }{(1+frac {1}{y})}^y=1$ @Physicist137 this limit is well known which is equal to 1.
                                $endgroup$
                                – Pentapolis
                                Aug 9 '16 at 22:09










                              • $begingroup$
                                Oh! Obviously. You are right. My mistake. Sorry... =). And +1. =).
                                $endgroup$
                                – Physicist137
                                Aug 10 '16 at 1:05




















                              • $begingroup$
                                The step from the third line to the last one is, to say the least, very questionable: how could you justify to take the limit of only a part of the expression in the right extreme of the line one before the last, but to leave other part still with $;y;$ ?
                                $endgroup$
                                – DonAntonio
                                Jul 27 '16 at 19:58










                              • $begingroup$
                                Yes, the step before the last step should be $={1^{frac {1}{1}}}$
                                $endgroup$
                                – Pentapolis
                                Jul 27 '16 at 20:50












                              • $begingroup$
                                If $yto 0$, then $1/ytoinfty$. Then in the end you have: $1+infty$, which is evidently not $1$.
                                $endgroup$
                                – Physicist137
                                Aug 9 '16 at 18:39












                              • $begingroup$
                                $lim _{yto 0 }{(1+frac {1}{y})}^y=1$ @Physicist137 this limit is well known which is equal to 1.
                                $endgroup$
                                – Pentapolis
                                Aug 9 '16 at 22:09










                              • $begingroup$
                                Oh! Obviously. You are right. My mistake. Sorry... =). And +1. =).
                                $endgroup$
                                – Physicist137
                                Aug 10 '16 at 1:05


















                              $begingroup$
                              The step from the third line to the last one is, to say the least, very questionable: how could you justify to take the limit of only a part of the expression in the right extreme of the line one before the last, but to leave other part still with $;y;$ ?
                              $endgroup$
                              – DonAntonio
                              Jul 27 '16 at 19:58




                              $begingroup$
                              The step from the third line to the last one is, to say the least, very questionable: how could you justify to take the limit of only a part of the expression in the right extreme of the line one before the last, but to leave other part still with $;y;$ ?
                              $endgroup$
                              – DonAntonio
                              Jul 27 '16 at 19:58












                              $begingroup$
                              Yes, the step before the last step should be $={1^{frac {1}{1}}}$
                              $endgroup$
                              – Pentapolis
                              Jul 27 '16 at 20:50






                              $begingroup$
                              Yes, the step before the last step should be $={1^{frac {1}{1}}}$
                              $endgroup$
                              – Pentapolis
                              Jul 27 '16 at 20:50














                              $begingroup$
                              If $yto 0$, then $1/ytoinfty$. Then in the end you have: $1+infty$, which is evidently not $1$.
                              $endgroup$
                              – Physicist137
                              Aug 9 '16 at 18:39






                              $begingroup$
                              If $yto 0$, then $1/ytoinfty$. Then in the end you have: $1+infty$, which is evidently not $1$.
                              $endgroup$
                              – Physicist137
                              Aug 9 '16 at 18:39














                              $begingroup$
                              $lim _{yto 0 }{(1+frac {1}{y})}^y=1$ @Physicist137 this limit is well known which is equal to 1.
                              $endgroup$
                              – Pentapolis
                              Aug 9 '16 at 22:09




                              $begingroup$
                              $lim _{yto 0 }{(1+frac {1}{y})}^y=1$ @Physicist137 this limit is well known which is equal to 1.
                              $endgroup$
                              – Pentapolis
                              Aug 9 '16 at 22:09












                              $begingroup$
                              Oh! Obviously. You are right. My mistake. Sorry... =). And +1. =).
                              $endgroup$
                              – Physicist137
                              Aug 10 '16 at 1:05






                              $begingroup$
                              Oh! Obviously. You are right. My mistake. Sorry... =). And +1. =).
                              $endgroup$
                              – Physicist137
                              Aug 10 '16 at 1:05




















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