Is there a way to find this limit algebraically? $limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}$












6












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I'm a Calculus I student and my teacher has given me a set of problems to solve with L'Hoptial's rule. Most of them have been pretty easy, but this one has me stumped.



$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}$$



You'll notice that using L'Hopital's rule flips the value of the top to the bottom. For example, using it once returns:



$$limlimits_{xto infty} frac{sqrt{x^2 + 1}}{x}$$



And doing it again returns you to the beginning:



$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}$$



I of course plugged it into my calculator to find the limit to evaluate to 1, but I was wondering if there was a better way to do this algebraically?










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$endgroup$








  • 6




    $begingroup$
    $frac{x}{sqrt{x^2 + 1}}=frac{sqrt{x^2}}{sqrt{x^2+1}}=sqrt{frac{x^2}{x^2+1}}$
    $endgroup$
    – Mason
    Jan 5 at 18:41






  • 3




    $begingroup$
    The "right" way to solve this is as Thomas Shelby wrote in his answer. However, you can almost intuitively see that the limit is $1$ even without using a calculator. Indeed, if $x$ is a huge positive number, say $10^{5863}$, the difference between $x^2$ and $x^2 + 1$ is extremely tiny. So the expression is essentially indistinguishable from ${xoversqrt{x^2}} = {xover x} = 1$.
    $endgroup$
    – Andreas Rejbrand
    Jan 5 at 19:37


















6












$begingroup$


I'm a Calculus I student and my teacher has given me a set of problems to solve with L'Hoptial's rule. Most of them have been pretty easy, but this one has me stumped.



$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}$$



You'll notice that using L'Hopital's rule flips the value of the top to the bottom. For example, using it once returns:



$$limlimits_{xto infty} frac{sqrt{x^2 + 1}}{x}$$



And doing it again returns you to the beginning:



$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}$$



I of course plugged it into my calculator to find the limit to evaluate to 1, but I was wondering if there was a better way to do this algebraically?










share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    $frac{x}{sqrt{x^2 + 1}}=frac{sqrt{x^2}}{sqrt{x^2+1}}=sqrt{frac{x^2}{x^2+1}}$
    $endgroup$
    – Mason
    Jan 5 at 18:41






  • 3




    $begingroup$
    The "right" way to solve this is as Thomas Shelby wrote in his answer. However, you can almost intuitively see that the limit is $1$ even without using a calculator. Indeed, if $x$ is a huge positive number, say $10^{5863}$, the difference between $x^2$ and $x^2 + 1$ is extremely tiny. So the expression is essentially indistinguishable from ${xoversqrt{x^2}} = {xover x} = 1$.
    $endgroup$
    – Andreas Rejbrand
    Jan 5 at 19:37
















6












6








6





$begingroup$


I'm a Calculus I student and my teacher has given me a set of problems to solve with L'Hoptial's rule. Most of them have been pretty easy, but this one has me stumped.



$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}$$



You'll notice that using L'Hopital's rule flips the value of the top to the bottom. For example, using it once returns:



$$limlimits_{xto infty} frac{sqrt{x^2 + 1}}{x}$$



And doing it again returns you to the beginning:



$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}$$



I of course plugged it into my calculator to find the limit to evaluate to 1, but I was wondering if there was a better way to do this algebraically?










share|cite|improve this question











$endgroup$




I'm a Calculus I student and my teacher has given me a set of problems to solve with L'Hoptial's rule. Most of them have been pretty easy, but this one has me stumped.



$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}$$



You'll notice that using L'Hopital's rule flips the value of the top to the bottom. For example, using it once returns:



$$limlimits_{xto infty} frac{sqrt{x^2 + 1}}{x}$$



And doing it again returns you to the beginning:



$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}$$



I of course plugged it into my calculator to find the limit to evaluate to 1, but I was wondering if there was a better way to do this algebraically?







calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 21:10









Blue

47.8k870152




47.8k870152










asked Jan 5 at 18:36









Jae SwanepoelJae Swanepoel

311




311








  • 6




    $begingroup$
    $frac{x}{sqrt{x^2 + 1}}=frac{sqrt{x^2}}{sqrt{x^2+1}}=sqrt{frac{x^2}{x^2+1}}$
    $endgroup$
    – Mason
    Jan 5 at 18:41






  • 3




    $begingroup$
    The "right" way to solve this is as Thomas Shelby wrote in his answer. However, you can almost intuitively see that the limit is $1$ even without using a calculator. Indeed, if $x$ is a huge positive number, say $10^{5863}$, the difference between $x^2$ and $x^2 + 1$ is extremely tiny. So the expression is essentially indistinguishable from ${xoversqrt{x^2}} = {xover x} = 1$.
    $endgroup$
    – Andreas Rejbrand
    Jan 5 at 19:37
















  • 6




    $begingroup$
    $frac{x}{sqrt{x^2 + 1}}=frac{sqrt{x^2}}{sqrt{x^2+1}}=sqrt{frac{x^2}{x^2+1}}$
    $endgroup$
    – Mason
    Jan 5 at 18:41






  • 3




    $begingroup$
    The "right" way to solve this is as Thomas Shelby wrote in his answer. However, you can almost intuitively see that the limit is $1$ even without using a calculator. Indeed, if $x$ is a huge positive number, say $10^{5863}$, the difference between $x^2$ and $x^2 + 1$ is extremely tiny. So the expression is essentially indistinguishable from ${xoversqrt{x^2}} = {xover x} = 1$.
    $endgroup$
    – Andreas Rejbrand
    Jan 5 at 19:37










6




6




$begingroup$
$frac{x}{sqrt{x^2 + 1}}=frac{sqrt{x^2}}{sqrt{x^2+1}}=sqrt{frac{x^2}{x^2+1}}$
$endgroup$
– Mason
Jan 5 at 18:41




$begingroup$
$frac{x}{sqrt{x^2 + 1}}=frac{sqrt{x^2}}{sqrt{x^2+1}}=sqrt{frac{x^2}{x^2+1}}$
$endgroup$
– Mason
Jan 5 at 18:41




3




3




$begingroup$
The "right" way to solve this is as Thomas Shelby wrote in his answer. However, you can almost intuitively see that the limit is $1$ even without using a calculator. Indeed, if $x$ is a huge positive number, say $10^{5863}$, the difference between $x^2$ and $x^2 + 1$ is extremely tiny. So the expression is essentially indistinguishable from ${xoversqrt{x^2}} = {xover x} = 1$.
$endgroup$
– Andreas Rejbrand
Jan 5 at 19:37






$begingroup$
The "right" way to solve this is as Thomas Shelby wrote in his answer. However, you can almost intuitively see that the limit is $1$ even without using a calculator. Indeed, if $x$ is a huge positive number, say $10^{5863}$, the difference between $x^2$ and $x^2 + 1$ is extremely tiny. So the expression is essentially indistinguishable from ${xoversqrt{x^2}} = {xover x} = 1$.
$endgroup$
– Andreas Rejbrand
Jan 5 at 19:37












5 Answers
5






active

oldest

votes


















13












$begingroup$

Hint: Divide the numerator and denominator by $x $ and apply the limit.



$$frac{x}{sqrt{x^2 + 1}}=frac{1}{sqrt{1 + frac{1}{x^2}}}$$






share|cite|improve this answer











$endgroup$









  • 6




    $begingroup$
    In other words, actually do the division you see written down in $lim_{xtoinfty}frac{sqrt{x^2+1}}x$.
    $endgroup$
    – Arthur
    Jan 5 at 18:40





















11












$begingroup$

Hint



Simply use $${xover x+1}={xover sqrt{x^2+2x+1}}<{xover sqrt{x^2+1}}<1$$for large enough $x>0$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Mostafa.Very nice+.
    $endgroup$
    – Peter Szilas
    Jan 5 at 19:01










  • $begingroup$
    But how so we know the limit is 1. Based on what you wrote couldnt the answer be 1/2
    $endgroup$
    – Milan Stojanovic
    Jan 5 at 19:05










  • $begingroup$
    @PeterSzilas thank you!
    $endgroup$
    – Mostafa Ayaz
    Jan 5 at 19:08






  • 3




    $begingroup$
    @MilanStojanovic No since $$lim_{xto infty}{xover x+1}=1$$and $$lim_{xto infty}1=1$$so the Squeeze theorem is applicable here....
    $endgroup$
    – Mostafa Ayaz
    Jan 5 at 19:09



















7












$begingroup$

By your own reasoning, you have the following:
$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=limlimits_{xto infty} frac{sqrt{x^2 + 1}}{x}$$



Now, the left side is clearly the reciprocal of the right side, so we have:
$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=frac{1}{limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}}$$



(Note that doing this manipulation assumes that $limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}$ converges to a real number. However, you can use the first derivative to show this is an always increasing function and then use basic algebra to show that $frac{x}{sqrt{x^2 + 1}} < 1$ for all $xinBbb{R}$. Thus, because this is a bounded, always increasing function, the limit as $xto infty$ must converge to some real number, so our assumption in this manipulation is valid.)



Cross-multiply:
$$left(limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}right)^2=1$$



Take the square root:
$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=pm 1$$



However, it is easy to show that $frac{x}{sqrt{x^2 + 1}}>0$ for all $x > 0$. Therefore, there's no way the limit can be a negative number like $-1$. Thus, the only possibility we have left is $+1$, so:
$$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=1$$






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    When computing the limit of rational functions, as is the case for $$lim_{x rightarrow infty} frac{x}{sqrt{x^2 +1}},$$ you want to divide the top and bottom by the highest degree in the denominator, which in this case is $x$. Since $x rightarrow +infty$, so $x$ is always positive (at least, near where we are worried about) I claim that $x = sqrt{x^2}$. So, if we divide the top and bottom by $x$, we get $$lim_{x rightarrow infty} frac{x}{sqrt{x^2 +1}} = lim_{x rightarrow infty} frac{1}{sqrt{1 + 1/x^2}}.$$ You should be able to compute the limit from here.



    Whenever you see a monomial in the numerator with the square root of a polynomial in the denominator, you should consider this method. Of course, keep in mind that you'll have to tweak it slightly if $x rightarrow -infty$! Try to see if you can figure out what would change in that case.






    share|cite|improve this answer











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      0












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      Set $x = sinh t$. We have
      $$frac{x}{sqrt{x^2+1}}= frac{sinh t}{sqrt{1+sinh^2t}} = frac{sinh t}{cosh t} = tanh t$$



      $x to infty$ is equivalent to $ttoinfty$ so $$lim_{xtoinfty} frac{x}{sqrt{x^2+1}} = lim_{ttoinfty} tanh t = lim_{ttoinfty}frac{e^t - e^{-t}}{e^t+e^{-t}} = lim_{ttoinfty}frac{e^{2t}-1}{e^{2t}+1} = 1$$






      share|cite|improve this answer









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        Your Answer





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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        13












        $begingroup$

        Hint: Divide the numerator and denominator by $x $ and apply the limit.



        $$frac{x}{sqrt{x^2 + 1}}=frac{1}{sqrt{1 + frac{1}{x^2}}}$$






        share|cite|improve this answer











        $endgroup$









        • 6




          $begingroup$
          In other words, actually do the division you see written down in $lim_{xtoinfty}frac{sqrt{x^2+1}}x$.
          $endgroup$
          – Arthur
          Jan 5 at 18:40


















        13












        $begingroup$

        Hint: Divide the numerator and denominator by $x $ and apply the limit.



        $$frac{x}{sqrt{x^2 + 1}}=frac{1}{sqrt{1 + frac{1}{x^2}}}$$






        share|cite|improve this answer











        $endgroup$









        • 6




          $begingroup$
          In other words, actually do the division you see written down in $lim_{xtoinfty}frac{sqrt{x^2+1}}x$.
          $endgroup$
          – Arthur
          Jan 5 at 18:40
















        13












        13








        13





        $begingroup$

        Hint: Divide the numerator and denominator by $x $ and apply the limit.



        $$frac{x}{sqrt{x^2 + 1}}=frac{1}{sqrt{1 + frac{1}{x^2}}}$$






        share|cite|improve this answer











        $endgroup$



        Hint: Divide the numerator and denominator by $x $ and apply the limit.



        $$frac{x}{sqrt{x^2 + 1}}=frac{1}{sqrt{1 + frac{1}{x^2}}}$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 5 at 18:46

























        answered Jan 5 at 18:38









        Thomas ShelbyThomas Shelby

        2,250220




        2,250220








        • 6




          $begingroup$
          In other words, actually do the division you see written down in $lim_{xtoinfty}frac{sqrt{x^2+1}}x$.
          $endgroup$
          – Arthur
          Jan 5 at 18:40
















        • 6




          $begingroup$
          In other words, actually do the division you see written down in $lim_{xtoinfty}frac{sqrt{x^2+1}}x$.
          $endgroup$
          – Arthur
          Jan 5 at 18:40










        6




        6




        $begingroup$
        In other words, actually do the division you see written down in $lim_{xtoinfty}frac{sqrt{x^2+1}}x$.
        $endgroup$
        – Arthur
        Jan 5 at 18:40






        $begingroup$
        In other words, actually do the division you see written down in $lim_{xtoinfty}frac{sqrt{x^2+1}}x$.
        $endgroup$
        – Arthur
        Jan 5 at 18:40













        11












        $begingroup$

        Hint



        Simply use $${xover x+1}={xover sqrt{x^2+2x+1}}<{xover sqrt{x^2+1}}<1$$for large enough $x>0$.






        share|cite|improve this answer









        $endgroup$









        • 1




          $begingroup$
          Mostafa.Very nice+.
          $endgroup$
          – Peter Szilas
          Jan 5 at 19:01










        • $begingroup$
          But how so we know the limit is 1. Based on what you wrote couldnt the answer be 1/2
          $endgroup$
          – Milan Stojanovic
          Jan 5 at 19:05










        • $begingroup$
          @PeterSzilas thank you!
          $endgroup$
          – Mostafa Ayaz
          Jan 5 at 19:08






        • 3




          $begingroup$
          @MilanStojanovic No since $$lim_{xto infty}{xover x+1}=1$$and $$lim_{xto infty}1=1$$so the Squeeze theorem is applicable here....
          $endgroup$
          – Mostafa Ayaz
          Jan 5 at 19:09
















        11












        $begingroup$

        Hint



        Simply use $${xover x+1}={xover sqrt{x^2+2x+1}}<{xover sqrt{x^2+1}}<1$$for large enough $x>0$.






        share|cite|improve this answer









        $endgroup$









        • 1




          $begingroup$
          Mostafa.Very nice+.
          $endgroup$
          – Peter Szilas
          Jan 5 at 19:01










        • $begingroup$
          But how so we know the limit is 1. Based on what you wrote couldnt the answer be 1/2
          $endgroup$
          – Milan Stojanovic
          Jan 5 at 19:05










        • $begingroup$
          @PeterSzilas thank you!
          $endgroup$
          – Mostafa Ayaz
          Jan 5 at 19:08






        • 3




          $begingroup$
          @MilanStojanovic No since $$lim_{xto infty}{xover x+1}=1$$and $$lim_{xto infty}1=1$$so the Squeeze theorem is applicable here....
          $endgroup$
          – Mostafa Ayaz
          Jan 5 at 19:09














        11












        11








        11





        $begingroup$

        Hint



        Simply use $${xover x+1}={xover sqrt{x^2+2x+1}}<{xover sqrt{x^2+1}}<1$$for large enough $x>0$.






        share|cite|improve this answer









        $endgroup$



        Hint



        Simply use $${xover x+1}={xover sqrt{x^2+2x+1}}<{xover sqrt{x^2+1}}<1$$for large enough $x>0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 18:53









        Mostafa AyazMostafa Ayaz

        15.2k3939




        15.2k3939








        • 1




          $begingroup$
          Mostafa.Very nice+.
          $endgroup$
          – Peter Szilas
          Jan 5 at 19:01










        • $begingroup$
          But how so we know the limit is 1. Based on what you wrote couldnt the answer be 1/2
          $endgroup$
          – Milan Stojanovic
          Jan 5 at 19:05










        • $begingroup$
          @PeterSzilas thank you!
          $endgroup$
          – Mostafa Ayaz
          Jan 5 at 19:08






        • 3




          $begingroup$
          @MilanStojanovic No since $$lim_{xto infty}{xover x+1}=1$$and $$lim_{xto infty}1=1$$so the Squeeze theorem is applicable here....
          $endgroup$
          – Mostafa Ayaz
          Jan 5 at 19:09














        • 1




          $begingroup$
          Mostafa.Very nice+.
          $endgroup$
          – Peter Szilas
          Jan 5 at 19:01










        • $begingroup$
          But how so we know the limit is 1. Based on what you wrote couldnt the answer be 1/2
          $endgroup$
          – Milan Stojanovic
          Jan 5 at 19:05










        • $begingroup$
          @PeterSzilas thank you!
          $endgroup$
          – Mostafa Ayaz
          Jan 5 at 19:08






        • 3




          $begingroup$
          @MilanStojanovic No since $$lim_{xto infty}{xover x+1}=1$$and $$lim_{xto infty}1=1$$so the Squeeze theorem is applicable here....
          $endgroup$
          – Mostafa Ayaz
          Jan 5 at 19:09








        1




        1




        $begingroup$
        Mostafa.Very nice+.
        $endgroup$
        – Peter Szilas
        Jan 5 at 19:01




        $begingroup$
        Mostafa.Very nice+.
        $endgroup$
        – Peter Szilas
        Jan 5 at 19:01












        $begingroup$
        But how so we know the limit is 1. Based on what you wrote couldnt the answer be 1/2
        $endgroup$
        – Milan Stojanovic
        Jan 5 at 19:05




        $begingroup$
        But how so we know the limit is 1. Based on what you wrote couldnt the answer be 1/2
        $endgroup$
        – Milan Stojanovic
        Jan 5 at 19:05












        $begingroup$
        @PeterSzilas thank you!
        $endgroup$
        – Mostafa Ayaz
        Jan 5 at 19:08




        $begingroup$
        @PeterSzilas thank you!
        $endgroup$
        – Mostafa Ayaz
        Jan 5 at 19:08




        3




        3




        $begingroup$
        @MilanStojanovic No since $$lim_{xto infty}{xover x+1}=1$$and $$lim_{xto infty}1=1$$so the Squeeze theorem is applicable here....
        $endgroup$
        – Mostafa Ayaz
        Jan 5 at 19:09




        $begingroup$
        @MilanStojanovic No since $$lim_{xto infty}{xover x+1}=1$$and $$lim_{xto infty}1=1$$so the Squeeze theorem is applicable here....
        $endgroup$
        – Mostafa Ayaz
        Jan 5 at 19:09











        7












        $begingroup$

        By your own reasoning, you have the following:
        $$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=limlimits_{xto infty} frac{sqrt{x^2 + 1}}{x}$$



        Now, the left side is clearly the reciprocal of the right side, so we have:
        $$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=frac{1}{limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}}$$



        (Note that doing this manipulation assumes that $limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}$ converges to a real number. However, you can use the first derivative to show this is an always increasing function and then use basic algebra to show that $frac{x}{sqrt{x^2 + 1}} < 1$ for all $xinBbb{R}$. Thus, because this is a bounded, always increasing function, the limit as $xto infty$ must converge to some real number, so our assumption in this manipulation is valid.)



        Cross-multiply:
        $$left(limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}right)^2=1$$



        Take the square root:
        $$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=pm 1$$



        However, it is easy to show that $frac{x}{sqrt{x^2 + 1}}>0$ for all $x > 0$. Therefore, there's no way the limit can be a negative number like $-1$. Thus, the only possibility we have left is $+1$, so:
        $$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=1$$






        share|cite|improve this answer









        $endgroup$


















          7












          $begingroup$

          By your own reasoning, you have the following:
          $$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=limlimits_{xto infty} frac{sqrt{x^2 + 1}}{x}$$



          Now, the left side is clearly the reciprocal of the right side, so we have:
          $$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=frac{1}{limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}}$$



          (Note that doing this manipulation assumes that $limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}$ converges to a real number. However, you can use the first derivative to show this is an always increasing function and then use basic algebra to show that $frac{x}{sqrt{x^2 + 1}} < 1$ for all $xinBbb{R}$. Thus, because this is a bounded, always increasing function, the limit as $xto infty$ must converge to some real number, so our assumption in this manipulation is valid.)



          Cross-multiply:
          $$left(limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}right)^2=1$$



          Take the square root:
          $$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=pm 1$$



          However, it is easy to show that $frac{x}{sqrt{x^2 + 1}}>0$ for all $x > 0$. Therefore, there's no way the limit can be a negative number like $-1$. Thus, the only possibility we have left is $+1$, so:
          $$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=1$$






          share|cite|improve this answer









          $endgroup$
















            7












            7








            7





            $begingroup$

            By your own reasoning, you have the following:
            $$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=limlimits_{xto infty} frac{sqrt{x^2 + 1}}{x}$$



            Now, the left side is clearly the reciprocal of the right side, so we have:
            $$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=frac{1}{limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}}$$



            (Note that doing this manipulation assumes that $limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}$ converges to a real number. However, you can use the first derivative to show this is an always increasing function and then use basic algebra to show that $frac{x}{sqrt{x^2 + 1}} < 1$ for all $xinBbb{R}$. Thus, because this is a bounded, always increasing function, the limit as $xto infty$ must converge to some real number, so our assumption in this manipulation is valid.)



            Cross-multiply:
            $$left(limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}right)^2=1$$



            Take the square root:
            $$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=pm 1$$



            However, it is easy to show that $frac{x}{sqrt{x^2 + 1}}>0$ for all $x > 0$. Therefore, there's no way the limit can be a negative number like $-1$. Thus, the only possibility we have left is $+1$, so:
            $$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=1$$






            share|cite|improve this answer









            $endgroup$



            By your own reasoning, you have the following:
            $$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=limlimits_{xto infty} frac{sqrt{x^2 + 1}}{x}$$



            Now, the left side is clearly the reciprocal of the right side, so we have:
            $$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=frac{1}{limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}}$$



            (Note that doing this manipulation assumes that $limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}$ converges to a real number. However, you can use the first derivative to show this is an always increasing function and then use basic algebra to show that $frac{x}{sqrt{x^2 + 1}} < 1$ for all $xinBbb{R}$. Thus, because this is a bounded, always increasing function, the limit as $xto infty$ must converge to some real number, so our assumption in this manipulation is valid.)



            Cross-multiply:
            $$left(limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}right)^2=1$$



            Take the square root:
            $$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=pm 1$$



            However, it is easy to show that $frac{x}{sqrt{x^2 + 1}}>0$ for all $x > 0$. Therefore, there's no way the limit can be a negative number like $-1$. Thus, the only possibility we have left is $+1$, so:
            $$limlimits_{xto infty} frac{x}{sqrt{x^2 + 1}}=1$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 5 at 18:46









            Noble MushtakNoble Mushtak

            15.2k1735




            15.2k1735























                4












                $begingroup$

                When computing the limit of rational functions, as is the case for $$lim_{x rightarrow infty} frac{x}{sqrt{x^2 +1}},$$ you want to divide the top and bottom by the highest degree in the denominator, which in this case is $x$. Since $x rightarrow +infty$, so $x$ is always positive (at least, near where we are worried about) I claim that $x = sqrt{x^2}$. So, if we divide the top and bottom by $x$, we get $$lim_{x rightarrow infty} frac{x}{sqrt{x^2 +1}} = lim_{x rightarrow infty} frac{1}{sqrt{1 + 1/x^2}}.$$ You should be able to compute the limit from here.



                Whenever you see a monomial in the numerator with the square root of a polynomial in the denominator, you should consider this method. Of course, keep in mind that you'll have to tweak it slightly if $x rightarrow -infty$! Try to see if you can figure out what would change in that case.






                share|cite|improve this answer











                $endgroup$


















                  4












                  $begingroup$

                  When computing the limit of rational functions, as is the case for $$lim_{x rightarrow infty} frac{x}{sqrt{x^2 +1}},$$ you want to divide the top and bottom by the highest degree in the denominator, which in this case is $x$. Since $x rightarrow +infty$, so $x$ is always positive (at least, near where we are worried about) I claim that $x = sqrt{x^2}$. So, if we divide the top and bottom by $x$, we get $$lim_{x rightarrow infty} frac{x}{sqrt{x^2 +1}} = lim_{x rightarrow infty} frac{1}{sqrt{1 + 1/x^2}}.$$ You should be able to compute the limit from here.



                  Whenever you see a monomial in the numerator with the square root of a polynomial in the denominator, you should consider this method. Of course, keep in mind that you'll have to tweak it slightly if $x rightarrow -infty$! Try to see if you can figure out what would change in that case.






                  share|cite|improve this answer











                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    When computing the limit of rational functions, as is the case for $$lim_{x rightarrow infty} frac{x}{sqrt{x^2 +1}},$$ you want to divide the top and bottom by the highest degree in the denominator, which in this case is $x$. Since $x rightarrow +infty$, so $x$ is always positive (at least, near where we are worried about) I claim that $x = sqrt{x^2}$. So, if we divide the top and bottom by $x$, we get $$lim_{x rightarrow infty} frac{x}{sqrt{x^2 +1}} = lim_{x rightarrow infty} frac{1}{sqrt{1 + 1/x^2}}.$$ You should be able to compute the limit from here.



                    Whenever you see a monomial in the numerator with the square root of a polynomial in the denominator, you should consider this method. Of course, keep in mind that you'll have to tweak it slightly if $x rightarrow -infty$! Try to see if you can figure out what would change in that case.






                    share|cite|improve this answer











                    $endgroup$



                    When computing the limit of rational functions, as is the case for $$lim_{x rightarrow infty} frac{x}{sqrt{x^2 +1}},$$ you want to divide the top and bottom by the highest degree in the denominator, which in this case is $x$. Since $x rightarrow +infty$, so $x$ is always positive (at least, near where we are worried about) I claim that $x = sqrt{x^2}$. So, if we divide the top and bottom by $x$, we get $$lim_{x rightarrow infty} frac{x}{sqrt{x^2 +1}} = lim_{x rightarrow infty} frac{1}{sqrt{1 + 1/x^2}}.$$ You should be able to compute the limit from here.



                    Whenever you see a monomial in the numerator with the square root of a polynomial in the denominator, you should consider this method. Of course, keep in mind that you'll have to tweak it slightly if $x rightarrow -infty$! Try to see if you can figure out what would change in that case.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 5 at 18:52









                    Noble Mushtak

                    15.2k1735




                    15.2k1735










                    answered Jan 5 at 18:51









                    kkckkc

                    1058




                    1058























                        0












                        $begingroup$

                        Set $x = sinh t$. We have
                        $$frac{x}{sqrt{x^2+1}}= frac{sinh t}{sqrt{1+sinh^2t}} = frac{sinh t}{cosh t} = tanh t$$



                        $x to infty$ is equivalent to $ttoinfty$ so $$lim_{xtoinfty} frac{x}{sqrt{x^2+1}} = lim_{ttoinfty} tanh t = lim_{ttoinfty}frac{e^t - e^{-t}}{e^t+e^{-t}} = lim_{ttoinfty}frac{e^{2t}-1}{e^{2t}+1} = 1$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Set $x = sinh t$. We have
                          $$frac{x}{sqrt{x^2+1}}= frac{sinh t}{sqrt{1+sinh^2t}} = frac{sinh t}{cosh t} = tanh t$$



                          $x to infty$ is equivalent to $ttoinfty$ so $$lim_{xtoinfty} frac{x}{sqrt{x^2+1}} = lim_{ttoinfty} tanh t = lim_{ttoinfty}frac{e^t - e^{-t}}{e^t+e^{-t}} = lim_{ttoinfty}frac{e^{2t}-1}{e^{2t}+1} = 1$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Set $x = sinh t$. We have
                            $$frac{x}{sqrt{x^2+1}}= frac{sinh t}{sqrt{1+sinh^2t}} = frac{sinh t}{cosh t} = tanh t$$



                            $x to infty$ is equivalent to $ttoinfty$ so $$lim_{xtoinfty} frac{x}{sqrt{x^2+1}} = lim_{ttoinfty} tanh t = lim_{ttoinfty}frac{e^t - e^{-t}}{e^t+e^{-t}} = lim_{ttoinfty}frac{e^{2t}-1}{e^{2t}+1} = 1$$






                            share|cite|improve this answer









                            $endgroup$



                            Set $x = sinh t$. We have
                            $$frac{x}{sqrt{x^2+1}}= frac{sinh t}{sqrt{1+sinh^2t}} = frac{sinh t}{cosh t} = tanh t$$



                            $x to infty$ is equivalent to $ttoinfty$ so $$lim_{xtoinfty} frac{x}{sqrt{x^2+1}} = lim_{ttoinfty} tanh t = lim_{ttoinfty}frac{e^t - e^{-t}}{e^t+e^{-t}} = lim_{ttoinfty}frac{e^{2t}-1}{e^{2t}+1} = 1$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 5 at 21:08









                            mechanodroidmechanodroid

                            27.1k62446




                            27.1k62446






























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