Solve limit $lim_{xto1}frac{x^2-1}{ln(x)}$ without using L'Hôpital's rule?












2












$begingroup$


enter image description here



How to solve this limit?



$$lim_{xto1}frac{x^2-1}{ln(x)}=frac00$$






calculus limits limits-without-lhopital







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  • $begingroup$
    Welcome to the site! Please see the MathJax guide to see how to format your math.
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    – Simply Beautiful Art
    Jan 14 '17 at 14:48












  • $begingroup$
    Not about the question, but that was a troll edit.
    $endgroup$
    – Cehhiro
    Jan 14 '17 at 14:49










  • $begingroup$
    Also, please do not mess up the pretty $LaTeX$...
    $endgroup$
    – Simply Beautiful Art
    Jan 14 '17 at 14:50
















2












$begingroup$


enter image description here



How to solve this limit?



$$lim_{xto1}frac{x^2-1}{ln(x)}=frac00$$






calculus limits limits-without-lhopital







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to the site! Please see the MathJax guide to see how to format your math.
    $endgroup$
    – Simply Beautiful Art
    Jan 14 '17 at 14:48












  • $begingroup$
    Not about the question, but that was a troll edit.
    $endgroup$
    – Cehhiro
    Jan 14 '17 at 14:49










  • $begingroup$
    Also, please do not mess up the pretty $LaTeX$...
    $endgroup$
    – Simply Beautiful Art
    Jan 14 '17 at 14:50














2












2








2





$begingroup$


enter image description here



How to solve this limit?



$$lim_{xto1}frac{x^2-1}{ln(x)}=frac00$$






calculus limits limits-without-lhopital







share|cite|improve this question











$endgroup$




enter image description here



How to solve this limit?



$$lim_{xto1}frac{x^2-1}{ln(x)}=frac00$$






calculus limits limits-without-lhopital




calculus 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 Jan 2 at 11:44









Nosrati

26.5k62354




26.5k62354










asked Jan 14 '17 at 14:47









Макс МотовиловМакс Мотовилов

92




92












  • $begingroup$
    Welcome to the site! Please see the MathJax guide to see how to format your math.
    $endgroup$
    – Simply Beautiful Art
    Jan 14 '17 at 14:48












  • $begingroup$
    Not about the question, but that was a troll edit.
    $endgroup$
    – Cehhiro
    Jan 14 '17 at 14:49










  • $begingroup$
    Also, please do not mess up the pretty $LaTeX$...
    $endgroup$
    – Simply Beautiful Art
    Jan 14 '17 at 14:50


















  • $begingroup$
    Welcome to the site! Please see the MathJax guide to see how to format your math.
    $endgroup$
    – Simply Beautiful Art
    Jan 14 '17 at 14:48












  • $begingroup$
    Not about the question, but that was a troll edit.
    $endgroup$
    – Cehhiro
    Jan 14 '17 at 14:49










  • $begingroup$
    Also, please do not mess up the pretty $LaTeX$...
    $endgroup$
    – Simply Beautiful Art
    Jan 14 '17 at 14:50
















$begingroup$
Welcome to the site! Please see the MathJax guide to see how to format your math.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:48






$begingroup$
Welcome to the site! Please see the MathJax guide to see how to format your math.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:48














$begingroup$
Not about the question, but that was a troll edit.
$endgroup$
– Cehhiro
Jan 14 '17 at 14:49




$begingroup$
Not about the question, but that was a troll edit.
$endgroup$
– Cehhiro
Jan 14 '17 at 14:49












$begingroup$
Also, please do not mess up the pretty $LaTeX$...
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:50




$begingroup$
Also, please do not mess up the pretty $LaTeX$...
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:50










5 Answers
5






active

oldest

votes


















2












$begingroup$

Set $x-1=hiff x=1+h$



$$lim_{xto1}dfrac{x^2-1}{ln x}=dfrac{lim_{hto0}dfrac{(h+1)^2-1}h}{lim_{hto0}dfrac{ln(1+h)}h}=?$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    :D That was pretty neat to see.
    $endgroup$
    – Simply Beautiful Art
    Jan 14 '17 at 14:53










  • $begingroup$
    Very neat, but doesn't he still need L'Hôpital for the denominator?
    $endgroup$
    – Cehhiro
    Jan 14 '17 at 14:54












  • $begingroup$
    @O.VonSeckendorff You need something to get past the $ln$, whether it be known limits or series expansion.
    $endgroup$
    – Simply Beautiful Art
    Jan 14 '17 at 14:56










  • $begingroup$
    @O.VonSeckendorff, This is an elementary formula of limit, readily available from $$lim_{xto0}dfrac{e^x-1}x=1$$ See math.stackexchange.com/questions/690898/…
    $endgroup$
    – lab bhattacharjee
    Jan 14 '17 at 14:56










  • $begingroup$
    @labbhattacharjee I didn't know/remember that one. Good to know.
    $endgroup$
    – Cehhiro
    Jan 14 '17 at 14:57



















2












$begingroup$

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities




$$bbox[5px,border:2px solid #C0A000]{frac{x-1}{x}le log(x)le x-1} tag 1$$




for $x>0$.



Hence, we have from $(1)$



$$frac{x^2-1}{x-1}le frac{x^2-1}{log(x)}le frac{x(x^2-1)}{x-1} tag 2$$



whence application of the squeeze theorem yields the coveted limit




$$lim_{xto 1}frac{x^2-1}{log(x)}=2$$







share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Plz. Check (1).
    $endgroup$
    – Nosrati
    Jan 14 '17 at 15:00










  • $begingroup$
    @MyGlasses Thank you; I've edited. -Mark
    $endgroup$
    – Mark Viola
    Jan 14 '17 at 15:04










  • $begingroup$
    Haha, nice... :-D
    $endgroup$
    – Simply Beautiful Art
    Jan 14 '17 at 15:09










  • $begingroup$
    @SimpleArt Thank you. And if you follow the link, it will show a way forward that begins with the limit definition of the exponential function.
    $endgroup$
    – Mark Viola
    Jan 14 '17 at 15:11



















1












$begingroup$

With substitution $x=e^t$ we have:
$$lim_{xto1}frac{x^2-1}{ln x}=lim_{tto0}frac{e^{2t}-1}{t}=lim_{tto0}frac{1+2t+frac{(2t)^2}{2}-1}{t}=2$$
where
$$e^theta=1+theta+frac{theta^2}{2}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    We can employ $$lim_{xto0}dfrac{e^x-1}x=1$$
    $endgroup$
    – lab bhattacharjee
    Jan 14 '17 at 15:00










  • $begingroup$
    @labbhattacharjee from where? without proof?
    $endgroup$
    – Nosrati
    Jan 14 '17 at 15:01










  • $begingroup$
    Which identities can be used for limit ? math.stackexchange.com/questions/42679/… OR 2000clicks.com/MathHelp/CalculusLimitExponential.aspx
    $endgroup$
    – lab bhattacharjee
    Jan 14 '17 at 15:04










  • $begingroup$
    I mean, it is limit definition to derivative of $e^x$ at $x=0$.
    $endgroup$
    – Simply Beautiful Art
    Jan 14 '17 at 15:09










  • $begingroup$
    @labbhattacharjee For using this limit, according to your links, all apply l'hopital rule, derivative definition and Taylor expansion. My suppose is the part of Taylor expansion also, but I don't know where can find a proof without these techniques.
    $endgroup$
    – Nosrati
    Jan 14 '17 at 15:09



















0












$begingroup$

Notice that



$$ln(x)=(x-1)-frac12(x-1)^2+mathcal O(x^3)$$



and



$$x^2-1=(x-1)(x+1)$$



So the reciprocal of the limit is



$$frac1L=lim_{xto1}frac1{x+1}-frac{(x-1)}{2(x+1)}+mathcal Oleft(frac{(x-1)^2}{x+1}right)=frac12$$



So that



$$L=2$$






share|cite|improve this answer









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    0












    $begingroup$

    By a shift of the variable your limit is equivalent to



    $$lim_{tto0}frac{t(t+1)}{ln(1+t)}=frac{lim_{tto0}t+1}{lim_{tto0}dfrac{ln(1+t)}t}=2.$$






    share|cite|improve this answer









    $endgroup$













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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Set $x-1=hiff x=1+h$



      $$lim_{xto1}dfrac{x^2-1}{ln x}=dfrac{lim_{hto0}dfrac{(h+1)^2-1}h}{lim_{hto0}dfrac{ln(1+h)}h}=?$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        :D That was pretty neat to see.
        $endgroup$
        – Simply Beautiful Art
        Jan 14 '17 at 14:53










      • $begingroup$
        Very neat, but doesn't he still need L'Hôpital for the denominator?
        $endgroup$
        – Cehhiro
        Jan 14 '17 at 14:54












      • $begingroup$
        @O.VonSeckendorff You need something to get past the $ln$, whether it be known limits or series expansion.
        $endgroup$
        – Simply Beautiful Art
        Jan 14 '17 at 14:56










      • $begingroup$
        @O.VonSeckendorff, This is an elementary formula of limit, readily available from $$lim_{xto0}dfrac{e^x-1}x=1$$ See math.stackexchange.com/questions/690898/…
        $endgroup$
        – lab bhattacharjee
        Jan 14 '17 at 14:56










      • $begingroup$
        @labbhattacharjee I didn't know/remember that one. Good to know.
        $endgroup$
        – Cehhiro
        Jan 14 '17 at 14:57
















      2












      $begingroup$

      Set $x-1=hiff x=1+h$



      $$lim_{xto1}dfrac{x^2-1}{ln x}=dfrac{lim_{hto0}dfrac{(h+1)^2-1}h}{lim_{hto0}dfrac{ln(1+h)}h}=?$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        :D That was pretty neat to see.
        $endgroup$
        – Simply Beautiful Art
        Jan 14 '17 at 14:53










      • $begingroup$
        Very neat, but doesn't he still need L'Hôpital for the denominator?
        $endgroup$
        – Cehhiro
        Jan 14 '17 at 14:54












      • $begingroup$
        @O.VonSeckendorff You need something to get past the $ln$, whether it be known limits or series expansion.
        $endgroup$
        – Simply Beautiful Art
        Jan 14 '17 at 14:56










      • $begingroup$
        @O.VonSeckendorff, This is an elementary formula of limit, readily available from $$lim_{xto0}dfrac{e^x-1}x=1$$ See math.stackexchange.com/questions/690898/…
        $endgroup$
        – lab bhattacharjee
        Jan 14 '17 at 14:56










      • $begingroup$
        @labbhattacharjee I didn't know/remember that one. Good to know.
        $endgroup$
        – Cehhiro
        Jan 14 '17 at 14:57














      2












      2








      2





      $begingroup$

      Set $x-1=hiff x=1+h$



      $$lim_{xto1}dfrac{x^2-1}{ln x}=dfrac{lim_{hto0}dfrac{(h+1)^2-1}h}{lim_{hto0}dfrac{ln(1+h)}h}=?$$






      share|cite|improve this answer











      $endgroup$



      Set $x-1=hiff x=1+h$



      $$lim_{xto1}dfrac{x^2-1}{ln x}=dfrac{lim_{hto0}dfrac{(h+1)^2-1}h}{lim_{hto0}dfrac{ln(1+h)}h}=?$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jan 14 '17 at 14:53

























      answered Jan 14 '17 at 14:52









      lab bhattacharjeelab bhattacharjee

      225k15156274




      225k15156274












      • $begingroup$
        :D That was pretty neat to see.
        $endgroup$
        – Simply Beautiful Art
        Jan 14 '17 at 14:53










      • $begingroup$
        Very neat, but doesn't he still need L'Hôpital for the denominator?
        $endgroup$
        – Cehhiro
        Jan 14 '17 at 14:54












      • $begingroup$
        @O.VonSeckendorff You need something to get past the $ln$, whether it be known limits or series expansion.
        $endgroup$
        – Simply Beautiful Art
        Jan 14 '17 at 14:56










      • $begingroup$
        @O.VonSeckendorff, This is an elementary formula of limit, readily available from $$lim_{xto0}dfrac{e^x-1}x=1$$ See math.stackexchange.com/questions/690898/…
        $endgroup$
        – lab bhattacharjee
        Jan 14 '17 at 14:56










      • $begingroup$
        @labbhattacharjee I didn't know/remember that one. Good to know.
        $endgroup$
        – Cehhiro
        Jan 14 '17 at 14:57


















      • $begingroup$
        :D That was pretty neat to see.
        $endgroup$
        – Simply Beautiful Art
        Jan 14 '17 at 14:53










      • $begingroup$
        Very neat, but doesn't he still need L'Hôpital for the denominator?
        $endgroup$
        – Cehhiro
        Jan 14 '17 at 14:54












      • $begingroup$
        @O.VonSeckendorff You need something to get past the $ln$, whether it be known limits or series expansion.
        $endgroup$
        – Simply Beautiful Art
        Jan 14 '17 at 14:56










      • $begingroup$
        @O.VonSeckendorff, This is an elementary formula of limit, readily available from $$lim_{xto0}dfrac{e^x-1}x=1$$ See math.stackexchange.com/questions/690898/…
        $endgroup$
        – lab bhattacharjee
        Jan 14 '17 at 14:56










      • $begingroup$
        @labbhattacharjee I didn't know/remember that one. Good to know.
        $endgroup$
        – Cehhiro
        Jan 14 '17 at 14:57
















      $begingroup$
      :D That was pretty neat to see.
      $endgroup$
      – Simply Beautiful Art
      Jan 14 '17 at 14:53




      $begingroup$
      :D That was pretty neat to see.
      $endgroup$
      – Simply Beautiful Art
      Jan 14 '17 at 14:53












      $begingroup$
      Very neat, but doesn't he still need L'Hôpital for the denominator?
      $endgroup$
      – Cehhiro
      Jan 14 '17 at 14:54






      $begingroup$
      Very neat, but doesn't he still need L'Hôpital for the denominator?
      $endgroup$
      – Cehhiro
      Jan 14 '17 at 14:54














      $begingroup$
      @O.VonSeckendorff You need something to get past the $ln$, whether it be known limits or series expansion.
      $endgroup$
      – Simply Beautiful Art
      Jan 14 '17 at 14:56




      $begingroup$
      @O.VonSeckendorff You need something to get past the $ln$, whether it be known limits or series expansion.
      $endgroup$
      – Simply Beautiful Art
      Jan 14 '17 at 14:56












      $begingroup$
      @O.VonSeckendorff, This is an elementary formula of limit, readily available from $$lim_{xto0}dfrac{e^x-1}x=1$$ See math.stackexchange.com/questions/690898/…
      $endgroup$
      – lab bhattacharjee
      Jan 14 '17 at 14:56




      $begingroup$
      @O.VonSeckendorff, This is an elementary formula of limit, readily available from $$lim_{xto0}dfrac{e^x-1}x=1$$ See math.stackexchange.com/questions/690898/…
      $endgroup$
      – lab bhattacharjee
      Jan 14 '17 at 14:56












      $begingroup$
      @labbhattacharjee I didn't know/remember that one. Good to know.
      $endgroup$
      – Cehhiro
      Jan 14 '17 at 14:57




      $begingroup$
      @labbhattacharjee I didn't know/remember that one. Good to know.
      $endgroup$
      – Cehhiro
      Jan 14 '17 at 14:57











      2












      $begingroup$

      In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities




      $$bbox[5px,border:2px solid #C0A000]{frac{x-1}{x}le log(x)le x-1} tag 1$$




      for $x>0$.



      Hence, we have from $(1)$



      $$frac{x^2-1}{x-1}le frac{x^2-1}{log(x)}le frac{x(x^2-1)}{x-1} tag 2$$



      whence application of the squeeze theorem yields the coveted limit




      $$lim_{xto 1}frac{x^2-1}{log(x)}=2$$







      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        Plz. Check (1).
        $endgroup$
        – Nosrati
        Jan 14 '17 at 15:00










      • $begingroup$
        @MyGlasses Thank you; I've edited. -Mark
        $endgroup$
        – Mark Viola
        Jan 14 '17 at 15:04










      • $begingroup$
        Haha, nice... :-D
        $endgroup$
        – Simply Beautiful Art
        Jan 14 '17 at 15:09










      • $begingroup$
        @SimpleArt Thank you. And if you follow the link, it will show a way forward that begins with the limit definition of the exponential function.
        $endgroup$
        – Mark Viola
        Jan 14 '17 at 15:11
















      2












      $begingroup$

      In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities




      $$bbox[5px,border:2px solid #C0A000]{frac{x-1}{x}le log(x)le x-1} tag 1$$




      for $x>0$.



      Hence, we have from $(1)$



      $$frac{x^2-1}{x-1}le frac{x^2-1}{log(x)}le frac{x(x^2-1)}{x-1} tag 2$$



      whence application of the squeeze theorem yields the coveted limit




      $$lim_{xto 1}frac{x^2-1}{log(x)}=2$$







      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        Plz. Check (1).
        $endgroup$
        – Nosrati
        Jan 14 '17 at 15:00










      • $begingroup$
        @MyGlasses Thank you; I've edited. -Mark
        $endgroup$
        – Mark Viola
        Jan 14 '17 at 15:04










      • $begingroup$
        Haha, nice... :-D
        $endgroup$
        – Simply Beautiful Art
        Jan 14 '17 at 15:09










      • $begingroup$
        @SimpleArt Thank you. And if you follow the link, it will show a way forward that begins with the limit definition of the exponential function.
        $endgroup$
        – Mark Viola
        Jan 14 '17 at 15:11














      2












      2








      2





      $begingroup$

      In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities




      $$bbox[5px,border:2px solid #C0A000]{frac{x-1}{x}le log(x)le x-1} tag 1$$




      for $x>0$.



      Hence, we have from $(1)$



      $$frac{x^2-1}{x-1}le frac{x^2-1}{log(x)}le frac{x(x^2-1)}{x-1} tag 2$$



      whence application of the squeeze theorem yields the coveted limit




      $$lim_{xto 1}frac{x^2-1}{log(x)}=2$$







      share|cite|improve this answer











      $endgroup$



      In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities




      $$bbox[5px,border:2px solid #C0A000]{frac{x-1}{x}le log(x)le x-1} tag 1$$




      for $x>0$.



      Hence, we have from $(1)$



      $$frac{x^2-1}{x-1}le frac{x^2-1}{log(x)}le frac{x(x^2-1)}{x-1} tag 2$$



      whence application of the squeeze theorem yields the coveted limit




      $$lim_{xto 1}frac{x^2-1}{log(x)}=2$$








      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Apr 13 '17 at 12:21









      Community♦

      1




      1










      answered Jan 14 '17 at 14:58









      Mark ViolaMark Viola

      131k1275171




      131k1275171








      • 1




        $begingroup$
        Plz. Check (1).
        $endgroup$
        – Nosrati
        Jan 14 '17 at 15:00










      • $begingroup$
        @MyGlasses Thank you; I've edited. -Mark
        $endgroup$
        – Mark Viola
        Jan 14 '17 at 15:04










      • $begingroup$
        Haha, nice... :-D
        $endgroup$
        – Simply Beautiful Art
        Jan 14 '17 at 15:09










      • $begingroup$
        @SimpleArt Thank you. And if you follow the link, it will show a way forward that begins with the limit definition of the exponential function.
        $endgroup$
        – Mark Viola
        Jan 14 '17 at 15:11














      • 1




        $begingroup$
        Plz. Check (1).
        $endgroup$
        – Nosrati
        Jan 14 '17 at 15:00










      • $begingroup$
        @MyGlasses Thank you; I've edited. -Mark
        $endgroup$
        – Mark Viola
        Jan 14 '17 at 15:04










      • $begingroup$
        Haha, nice... :-D
        $endgroup$
        – Simply Beautiful Art
        Jan 14 '17 at 15:09










      • $begingroup$
        @SimpleArt Thank you. And if you follow the link, it will show a way forward that begins with the limit definition of the exponential function.
        $endgroup$
        – Mark Viola
        Jan 14 '17 at 15:11








      1




      1




      $begingroup$
      Plz. Check (1).
      $endgroup$
      – Nosrati
      Jan 14 '17 at 15:00




      $begingroup$
      Plz. Check (1).
      $endgroup$
      – Nosrati
      Jan 14 '17 at 15:00












      $begingroup$
      @MyGlasses Thank you; I've edited. -Mark
      $endgroup$
      – Mark Viola
      Jan 14 '17 at 15:04




      $begingroup$
      @MyGlasses Thank you; I've edited. -Mark
      $endgroup$
      – Mark Viola
      Jan 14 '17 at 15:04












      $begingroup$
      Haha, nice... :-D
      $endgroup$
      – Simply Beautiful Art
      Jan 14 '17 at 15:09




      $begingroup$
      Haha, nice... :-D
      $endgroup$
      – Simply Beautiful Art
      Jan 14 '17 at 15:09












      $begingroup$
      @SimpleArt Thank you. And if you follow the link, it will show a way forward that begins with the limit definition of the exponential function.
      $endgroup$
      – Mark Viola
      Jan 14 '17 at 15:11




      $begingroup$
      @SimpleArt Thank you. And if you follow the link, it will show a way forward that begins with the limit definition of the exponential function.
      $endgroup$
      – Mark Viola
      Jan 14 '17 at 15:11











      1












      $begingroup$

      With substitution $x=e^t$ we have:
      $$lim_{xto1}frac{x^2-1}{ln x}=lim_{tto0}frac{e^{2t}-1}{t}=lim_{tto0}frac{1+2t+frac{(2t)^2}{2}-1}{t}=2$$
      where
      $$e^theta=1+theta+frac{theta^2}{2}$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        We can employ $$lim_{xto0}dfrac{e^x-1}x=1$$
        $endgroup$
        – lab bhattacharjee
        Jan 14 '17 at 15:00










      • $begingroup$
        @labbhattacharjee from where? without proof?
        $endgroup$
        – Nosrati
        Jan 14 '17 at 15:01










      • $begingroup$
        Which identities can be used for limit ? math.stackexchange.com/questions/42679/… OR 2000clicks.com/MathHelp/CalculusLimitExponential.aspx
        $endgroup$
        – lab bhattacharjee
        Jan 14 '17 at 15:04










      • $begingroup$
        I mean, it is limit definition to derivative of $e^x$ at $x=0$.
        $endgroup$
        – Simply Beautiful Art
        Jan 14 '17 at 15:09










      • $begingroup$
        @labbhattacharjee For using this limit, according to your links, all apply l'hopital rule, derivative definition and Taylor expansion. My suppose is the part of Taylor expansion also, but I don't know where can find a proof without these techniques.
        $endgroup$
        – Nosrati
        Jan 14 '17 at 15:09
















      1












      $begingroup$

      With substitution $x=e^t$ we have:
      $$lim_{xto1}frac{x^2-1}{ln x}=lim_{tto0}frac{e^{2t}-1}{t}=lim_{tto0}frac{1+2t+frac{(2t)^2}{2}-1}{t}=2$$
      where
      $$e^theta=1+theta+frac{theta^2}{2}$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        We can employ $$lim_{xto0}dfrac{e^x-1}x=1$$
        $endgroup$
        – lab bhattacharjee
        Jan 14 '17 at 15:00










      • $begingroup$
        @labbhattacharjee from where? without proof?
        $endgroup$
        – Nosrati
        Jan 14 '17 at 15:01










      • $begingroup$
        Which identities can be used for limit ? math.stackexchange.com/questions/42679/… OR 2000clicks.com/MathHelp/CalculusLimitExponential.aspx
        $endgroup$
        – lab bhattacharjee
        Jan 14 '17 at 15:04










      • $begingroup$
        I mean, it is limit definition to derivative of $e^x$ at $x=0$.
        $endgroup$
        – Simply Beautiful Art
        Jan 14 '17 at 15:09










      • $begingroup$
        @labbhattacharjee For using this limit, according to your links, all apply l'hopital rule, derivative definition and Taylor expansion. My suppose is the part of Taylor expansion also, but I don't know where can find a proof without these techniques.
        $endgroup$
        – Nosrati
        Jan 14 '17 at 15:09














      1












      1








      1





      $begingroup$

      With substitution $x=e^t$ we have:
      $$lim_{xto1}frac{x^2-1}{ln x}=lim_{tto0}frac{e^{2t}-1}{t}=lim_{tto0}frac{1+2t+frac{(2t)^2}{2}-1}{t}=2$$
      where
      $$e^theta=1+theta+frac{theta^2}{2}$$






      share|cite|improve this answer











      $endgroup$



      With substitution $x=e^t$ we have:
      $$lim_{xto1}frac{x^2-1}{ln x}=lim_{tto0}frac{e^{2t}-1}{t}=lim_{tto0}frac{1+2t+frac{(2t)^2}{2}-1}{t}=2$$
      where
      $$e^theta=1+theta+frac{theta^2}{2}$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jan 14 '17 at 15:04

























      answered Jan 14 '17 at 14:58









      NosratiNosrati

      26.5k62354




      26.5k62354












      • $begingroup$
        We can employ $$lim_{xto0}dfrac{e^x-1}x=1$$
        $endgroup$
        – lab bhattacharjee
        Jan 14 '17 at 15:00










      • $begingroup$
        @labbhattacharjee from where? without proof?
        $endgroup$
        – Nosrati
        Jan 14 '17 at 15:01










      • $begingroup$
        Which identities can be used for limit ? math.stackexchange.com/questions/42679/… OR 2000clicks.com/MathHelp/CalculusLimitExponential.aspx
        $endgroup$
        – lab bhattacharjee
        Jan 14 '17 at 15:04










      • $begingroup$
        I mean, it is limit definition to derivative of $e^x$ at $x=0$.
        $endgroup$
        – Simply Beautiful Art
        Jan 14 '17 at 15:09










      • $begingroup$
        @labbhattacharjee For using this limit, according to your links, all apply l'hopital rule, derivative definition and Taylor expansion. My suppose is the part of Taylor expansion also, but I don't know where can find a proof without these techniques.
        $endgroup$
        – Nosrati
        Jan 14 '17 at 15:09


















      • $begingroup$
        We can employ $$lim_{xto0}dfrac{e^x-1}x=1$$
        $endgroup$
        – lab bhattacharjee
        Jan 14 '17 at 15:00










      • $begingroup$
        @labbhattacharjee from where? without proof?
        $endgroup$
        – Nosrati
        Jan 14 '17 at 15:01










      • $begingroup$
        Which identities can be used for limit ? math.stackexchange.com/questions/42679/… OR 2000clicks.com/MathHelp/CalculusLimitExponential.aspx
        $endgroup$
        – lab bhattacharjee
        Jan 14 '17 at 15:04










      • $begingroup$
        I mean, it is limit definition to derivative of $e^x$ at $x=0$.
        $endgroup$
        – Simply Beautiful Art
        Jan 14 '17 at 15:09










      • $begingroup$
        @labbhattacharjee For using this limit, according to your links, all apply l'hopital rule, derivative definition and Taylor expansion. My suppose is the part of Taylor expansion also, but I don't know where can find a proof without these techniques.
        $endgroup$
        – Nosrati
        Jan 14 '17 at 15:09
















      $begingroup$
      We can employ $$lim_{xto0}dfrac{e^x-1}x=1$$
      $endgroup$
      – lab bhattacharjee
      Jan 14 '17 at 15:00




      $begingroup$
      We can employ $$lim_{xto0}dfrac{e^x-1}x=1$$
      $endgroup$
      – lab bhattacharjee
      Jan 14 '17 at 15:00












      $begingroup$
      @labbhattacharjee from where? without proof?
      $endgroup$
      – Nosrati
      Jan 14 '17 at 15:01




      $begingroup$
      @labbhattacharjee from where? without proof?
      $endgroup$
      – Nosrati
      Jan 14 '17 at 15:01












      $begingroup$
      Which identities can be used for limit ? math.stackexchange.com/questions/42679/… OR 2000clicks.com/MathHelp/CalculusLimitExponential.aspx
      $endgroup$
      – lab bhattacharjee
      Jan 14 '17 at 15:04




      $begingroup$
      Which identities can be used for limit ? math.stackexchange.com/questions/42679/… OR 2000clicks.com/MathHelp/CalculusLimitExponential.aspx
      $endgroup$
      – lab bhattacharjee
      Jan 14 '17 at 15:04












      $begingroup$
      I mean, it is limit definition to derivative of $e^x$ at $x=0$.
      $endgroup$
      – Simply Beautiful Art
      Jan 14 '17 at 15:09




      $begingroup$
      I mean, it is limit definition to derivative of $e^x$ at $x=0$.
      $endgroup$
      – Simply Beautiful Art
      Jan 14 '17 at 15:09












      $begingroup$
      @labbhattacharjee For using this limit, according to your links, all apply l'hopital rule, derivative definition and Taylor expansion. My suppose is the part of Taylor expansion also, but I don't know where can find a proof without these techniques.
      $endgroup$
      – Nosrati
      Jan 14 '17 at 15:09




      $begingroup$
      @labbhattacharjee For using this limit, according to your links, all apply l'hopital rule, derivative definition and Taylor expansion. My suppose is the part of Taylor expansion also, but I don't know where can find a proof without these techniques.
      $endgroup$
      – Nosrati
      Jan 14 '17 at 15:09











      0












      $begingroup$

      Notice that



      $$ln(x)=(x-1)-frac12(x-1)^2+mathcal O(x^3)$$



      and



      $$x^2-1=(x-1)(x+1)$$



      So the reciprocal of the limit is



      $$frac1L=lim_{xto1}frac1{x+1}-frac{(x-1)}{2(x+1)}+mathcal Oleft(frac{(x-1)^2}{x+1}right)=frac12$$



      So that



      $$L=2$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Notice that



        $$ln(x)=(x-1)-frac12(x-1)^2+mathcal O(x^3)$$



        and



        $$x^2-1=(x-1)(x+1)$$



        So the reciprocal of the limit is



        $$frac1L=lim_{xto1}frac1{x+1}-frac{(x-1)}{2(x+1)}+mathcal Oleft(frac{(x-1)^2}{x+1}right)=frac12$$



        So that



        $$L=2$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Notice that



          $$ln(x)=(x-1)-frac12(x-1)^2+mathcal O(x^3)$$



          and



          $$x^2-1=(x-1)(x+1)$$



          So the reciprocal of the limit is



          $$frac1L=lim_{xto1}frac1{x+1}-frac{(x-1)}{2(x+1)}+mathcal Oleft(frac{(x-1)^2}{x+1}right)=frac12$$



          So that



          $$L=2$$






          share|cite|improve this answer









          $endgroup$



          Notice that



          $$ln(x)=(x-1)-frac12(x-1)^2+mathcal O(x^3)$$



          and



          $$x^2-1=(x-1)(x+1)$$



          So the reciprocal of the limit is



          $$frac1L=lim_{xto1}frac1{x+1}-frac{(x-1)}{2(x+1)}+mathcal Oleft(frac{(x-1)^2}{x+1}right)=frac12$$



          So that



          $$L=2$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 14 '17 at 14:52









          Simply Beautiful ArtSimply Beautiful Art

          50.5k578181




          50.5k578181























              0












              $begingroup$

              By a shift of the variable your limit is equivalent to



              $$lim_{tto0}frac{t(t+1)}{ln(1+t)}=frac{lim_{tto0}t+1}{lim_{tto0}dfrac{ln(1+t)}t}=2.$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                By a shift of the variable your limit is equivalent to



                $$lim_{tto0}frac{t(t+1)}{ln(1+t)}=frac{lim_{tto0}t+1}{lim_{tto0}dfrac{ln(1+t)}t}=2.$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  By a shift of the variable your limit is equivalent to



                  $$lim_{tto0}frac{t(t+1)}{ln(1+t)}=frac{lim_{tto0}t+1}{lim_{tto0}dfrac{ln(1+t)}t}=2.$$






                  share|cite|improve this answer









                  $endgroup$



                  By a shift of the variable your limit is equivalent to



                  $$lim_{tto0}frac{t(t+1)}{ln(1+t)}=frac{lim_{tto0}t+1}{lim_{tto0}dfrac{ln(1+t)}t}=2.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 14 '17 at 15:32









                  Yves DaoustYves Daoust

                  125k671223




                  125k671223






























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