How to find the derivative of the function at $0$












1















How to find the derivative of the function at $0$



$f(x)=e^{-frac{1}{x^2}}sinfrac{1}{x}$ if $xne0$



$=0$ if $x=0$




I cannnot evaluate the limit $$lim_{xto0}frac{e^{-frac{1}{x^2}}sinfrac{1}{x}}{x}.$$



L'Hospital is not working here.



Please help.






real-analysis







share|cite|improve this question



























    1















    How to find the derivative of the function at $0$



    $f(x)=e^{-frac{1}{x^2}}sinfrac{1}{x}$ if $xne0$



    $=0$ if $x=0$




    I cannnot evaluate the limit $$lim_{xto0}frac{e^{-frac{1}{x^2}}sinfrac{1}{x}}{x}.$$



    L'Hospital is not working here.



    Please help.






    real-analysis







    share|cite|improve this question

























      1












      1








      1








      How to find the derivative of the function at $0$



      $f(x)=e^{-frac{1}{x^2}}sinfrac{1}{x}$ if $xne0$



      $=0$ if $x=0$




      I cannnot evaluate the limit $$lim_{xto0}frac{e^{-frac{1}{x^2}}sinfrac{1}{x}}{x}.$$



      L'Hospital is not working here.



      Please help.






      real-analysis







      share|cite|improve this question














      How to find the derivative of the function at $0$



      $f(x)=e^{-frac{1}{x^2}}sinfrac{1}{x}$ if $xne0$



      $=0$ if $x=0$




      I cannnot evaluate the limit $$lim_{xto0}frac{e^{-frac{1}{x^2}}sinfrac{1}{x}}{x}.$$



      L'Hospital is not working here.



      Please help.






      real-analysis




      real-analysis






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 22 at 3:11









      Jave

      458114




      458114






















          3 Answers
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          4














          Hint 1: $-1le sin(1/x)le 1$.



          Hint 2: $displaystyle lim_{xto 0}frac{1}{x}e^{-1/x^2} = lim_{yto pminfty} ycdot e^{-y^2}$.






          share|cite|improve this answer























          • I have tried this but could not solve yet the rest of the limit.
            – Jave
            Nov 22 at 3:25



















          2














          You already know that $|sin(1/x)|le 1$. This helps you only if the rest is $0$. So do a change of variable $y=1/x$ and use $e^{-frac 1{x^2}}=frac 1 {e^{y^2}}$. Now you need to calculate $$lim_{ytopminfty}frac y{e^{y^2}}$$






          share|cite|improve this answer





























            1














            $-1le sin(1/x)le 1 ::text{and let $frac1x=t$}implies -dfrac{t}{e^{t^2}}leq sin tdfrac{t} {e^{t^2}} leq dfrac{t}{e^{t^2}}$ Taking limit as $trightarrow infty$ and using L'hopital rule we get $displaystyle lim_{trightarrowinfty}sin tdfrac{t} {e^{t^2}}=0$.






            share|cite|improve this answer





















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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              4














              Hint 1: $-1le sin(1/x)le 1$.



              Hint 2: $displaystyle lim_{xto 0}frac{1}{x}e^{-1/x^2} = lim_{yto pminfty} ycdot e^{-y^2}$.






              share|cite|improve this answer























              • I have tried this but could not solve yet the rest of the limit.
                – Jave
                Nov 22 at 3:25
















              4














              Hint 1: $-1le sin(1/x)le 1$.



              Hint 2: $displaystyle lim_{xto 0}frac{1}{x}e^{-1/x^2} = lim_{yto pminfty} ycdot e^{-y^2}$.






              share|cite|improve this answer























              • I have tried this but could not solve yet the rest of the limit.
                – Jave
                Nov 22 at 3:25














              4












              4








              4






              Hint 1: $-1le sin(1/x)le 1$.



              Hint 2: $displaystyle lim_{xto 0}frac{1}{x}e^{-1/x^2} = lim_{yto pminfty} ycdot e^{-y^2}$.






              share|cite|improve this answer














              Hint 1: $-1le sin(1/x)le 1$.



              Hint 2: $displaystyle lim_{xto 0}frac{1}{x}e^{-1/x^2} = lim_{yto pminfty} ycdot e^{-y^2}$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 22 at 3:36

























              answered Nov 22 at 3:23









              Torsten Schoeneberg

              3,6812833




              3,6812833












              • I have tried this but could not solve yet the rest of the limit.
                – Jave
                Nov 22 at 3:25


















              • I have tried this but could not solve yet the rest of the limit.
                – Jave
                Nov 22 at 3:25
















              I have tried this but could not solve yet the rest of the limit.
              – Jave
              Nov 22 at 3:25




              I have tried this but could not solve yet the rest of the limit.
              – Jave
              Nov 22 at 3:25











              2














              You already know that $|sin(1/x)|le 1$. This helps you only if the rest is $0$. So do a change of variable $y=1/x$ and use $e^{-frac 1{x^2}}=frac 1 {e^{y^2}}$. Now you need to calculate $$lim_{ytopminfty}frac y{e^{y^2}}$$






              share|cite|improve this answer


























                2














                You already know that $|sin(1/x)|le 1$. This helps you only if the rest is $0$. So do a change of variable $y=1/x$ and use $e^{-frac 1{x^2}}=frac 1 {e^{y^2}}$. Now you need to calculate $$lim_{ytopminfty}frac y{e^{y^2}}$$






                share|cite|improve this answer
























                  2












                  2








                  2






                  You already know that $|sin(1/x)|le 1$. This helps you only if the rest is $0$. So do a change of variable $y=1/x$ and use $e^{-frac 1{x^2}}=frac 1 {e^{y^2}}$. Now you need to calculate $$lim_{ytopminfty}frac y{e^{y^2}}$$






                  share|cite|improve this answer












                  You already know that $|sin(1/x)|le 1$. This helps you only if the rest is $0$. So do a change of variable $y=1/x$ and use $e^{-frac 1{x^2}}=frac 1 {e^{y^2}}$. Now you need to calculate $$lim_{ytopminfty}frac y{e^{y^2}}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 at 3:34









                  Andrei

                  10.8k21025




                  10.8k21025























                      1














                      $-1le sin(1/x)le 1 ::text{and let $frac1x=t$}implies -dfrac{t}{e^{t^2}}leq sin tdfrac{t} {e^{t^2}} leq dfrac{t}{e^{t^2}}$ Taking limit as $trightarrow infty$ and using L'hopital rule we get $displaystyle lim_{trightarrowinfty}sin tdfrac{t} {e^{t^2}}=0$.






                      share|cite|improve this answer


























                        1














                        $-1le sin(1/x)le 1 ::text{and let $frac1x=t$}implies -dfrac{t}{e^{t^2}}leq sin tdfrac{t} {e^{t^2}} leq dfrac{t}{e^{t^2}}$ Taking limit as $trightarrow infty$ and using L'hopital rule we get $displaystyle lim_{trightarrowinfty}sin tdfrac{t} {e^{t^2}}=0$.






                        share|cite|improve this answer
























                          1












                          1








                          1






                          $-1le sin(1/x)le 1 ::text{and let $frac1x=t$}implies -dfrac{t}{e^{t^2}}leq sin tdfrac{t} {e^{t^2}} leq dfrac{t}{e^{t^2}}$ Taking limit as $trightarrow infty$ and using L'hopital rule we get $displaystyle lim_{trightarrowinfty}sin tdfrac{t} {e^{t^2}}=0$.






                          share|cite|improve this answer












                          $-1le sin(1/x)le 1 ::text{and let $frac1x=t$}implies -dfrac{t}{e^{t^2}}leq sin tdfrac{t} {e^{t^2}} leq dfrac{t}{e^{t^2}}$ Taking limit as $trightarrow infty$ and using L'hopital rule we get $displaystyle lim_{trightarrowinfty}sin tdfrac{t} {e^{t^2}}=0$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 22 at 3:40









                          Yadati Kiran

                          1,698519




                          1,698519






























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