Showing that a function is differentiable at $0$ with $f'(0)=0$












2












$begingroup$


The function is defined by



$$f(x) = begin{cases} e^{-1/x^{2}} & text{if $x neq 0$} \
0 & text{if $x = 0$} end{cases} $$



I tried using this definition



$$f'(a) = lim_{x to a} frac {f(x)-f(a)}{x-a}$$



which results in having



$$lim_{x to 0} frac {e^{-1/x^{2}}-0}{x-0} = lim_{x to 0} frac {e^{-1/x^{2}}}{x}$$



I see that you can solve this using L'Hospital's rules, but the value is still $frac 00$ no matter how many times I do the differential to each of the fraction.



What is the best way to approach this?






real-analysis







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    The function is defined by



    $$f(x) = begin{cases} e^{-1/x^{2}} & text{if $x neq 0$} \
    0 & text{if $x = 0$} end{cases} $$



    I tried using this definition



    $$f'(a) = lim_{x to a} frac {f(x)-f(a)}{x-a}$$



    which results in having



    $$lim_{x to 0} frac {e^{-1/x^{2}}-0}{x-0} = lim_{x to 0} frac {e^{-1/x^{2}}}{x}$$



    I see that you can solve this using L'Hospital's rules, but the value is still $frac 00$ no matter how many times I do the differential to each of the fraction.



    What is the best way to approach this?






    real-analysis







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      The function is defined by



      $$f(x) = begin{cases} e^{-1/x^{2}} & text{if $x neq 0$} \
      0 & text{if $x = 0$} end{cases} $$



      I tried using this definition



      $$f'(a) = lim_{x to a} frac {f(x)-f(a)}{x-a}$$



      which results in having



      $$lim_{x to 0} frac {e^{-1/x^{2}}-0}{x-0} = lim_{x to 0} frac {e^{-1/x^{2}}}{x}$$



      I see that you can solve this using L'Hospital's rules, but the value is still $frac 00$ no matter how many times I do the differential to each of the fraction.



      What is the best way to approach this?






      real-analysis







      share|cite|improve this question











      $endgroup$




      The function is defined by



      $$f(x) = begin{cases} e^{-1/x^{2}} & text{if $x neq 0$} \
      0 & text{if $x = 0$} end{cases} $$



      I tried using this definition



      $$f'(a) = lim_{x to a} frac {f(x)-f(a)}{x-a}$$



      which results in having



      $$lim_{x to 0} frac {e^{-1/x^{2}}-0}{x-0} = lim_{x to 0} frac {e^{-1/x^{2}}}{x}$$



      I see that you can solve this using L'Hospital's rules, but the value is still $frac 00$ no matter how many times I do the differential to each of the fraction.



      What is the best way to approach this?






      real-analysis




      real-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 7 '18 at 7:06









      Kemono Chen

      3,0521743




      3,0521743










      asked Dec 7 '18 at 5:58









      Jeffry SantosaJeffry Santosa

      384




      384






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Hint
          $$lim_{xto0}frac{e^{-1/x^2}}x=lim_{xtopminfty}e^{-x^2}{x}=lim_{xtopminfty}frac{x}{e^{x^2}}=lim_{xtopminfty}frac{1}{2xe^{x^2}}$$
          Explaination
          $$lim_{xto0}f(x)=lim_{xtopminfty}f(1/x)$$
          It can be easily proved by using epsilon-delta language.

          I applied L'hospital in the last equal sign.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm sorry this may sound embarrassing, even though I'm a 3rd year math major. Can you elaborate more on your hint? How does $lim_{xto0}frac{e^{-1/x^2}}x$ becomes $lim_{xtoinfty}e^{-x^2}{x}$?
            $endgroup$
            – Jeffry Santosa
            Dec 7 '18 at 6:12












          • $begingroup$
            If you need more explanation, just ask.
            $endgroup$
            – Kemono Chen
            Dec 7 '18 at 6:17










          • $begingroup$
            Note that $lim_{xto 0} f(x) = lim_{x to infty} f(1/x)$ is not true in general. If the function is not continuous at zero the latter limit may still exist.
            $endgroup$
            – Pjotr5
            Dec 7 '18 at 8:09










          • $begingroup$
            @Pjotr5 Here $xtoinfty$ denotes $xtopminfty$.
            $endgroup$
            – Kemono Chen
            Dec 7 '18 at 8:10












          • $begingroup$
            @KemonoChen To be very precise it would be good to mention this, because that is not standard notation for real limits.
            $endgroup$
            – Pjotr5
            Dec 7 '18 at 8:19



















          1












          $begingroup$

          We have
          $$lim_{xrightarrow 0}frac{e^{-1/x^2}}{x}=lim_{trightarrow infty}frac{t}{e^{t^2}}=lim_{trightarrow infty}frac{1}{2te^{t^2}}=0 $$
          by $t=frac{1}{x}$ and L'Hospital's rules.






          share|cite|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Hint
            $$lim_{xto0}frac{e^{-1/x^2}}x=lim_{xtopminfty}e^{-x^2}{x}=lim_{xtopminfty}frac{x}{e^{x^2}}=lim_{xtopminfty}frac{1}{2xe^{x^2}}$$
            Explaination
            $$lim_{xto0}f(x)=lim_{xtopminfty}f(1/x)$$
            It can be easily proved by using epsilon-delta language.

            I applied L'hospital in the last equal sign.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I'm sorry this may sound embarrassing, even though I'm a 3rd year math major. Can you elaborate more on your hint? How does $lim_{xto0}frac{e^{-1/x^2}}x$ becomes $lim_{xtoinfty}e^{-x^2}{x}$?
              $endgroup$
              – Jeffry Santosa
              Dec 7 '18 at 6:12












            • $begingroup$
              If you need more explanation, just ask.
              $endgroup$
              – Kemono Chen
              Dec 7 '18 at 6:17










            • $begingroup$
              Note that $lim_{xto 0} f(x) = lim_{x to infty} f(1/x)$ is not true in general. If the function is not continuous at zero the latter limit may still exist.
              $endgroup$
              – Pjotr5
              Dec 7 '18 at 8:09










            • $begingroup$
              @Pjotr5 Here $xtoinfty$ denotes $xtopminfty$.
              $endgroup$
              – Kemono Chen
              Dec 7 '18 at 8:10












            • $begingroup$
              @KemonoChen To be very precise it would be good to mention this, because that is not standard notation for real limits.
              $endgroup$
              – Pjotr5
              Dec 7 '18 at 8:19
















            1












            $begingroup$

            Hint
            $$lim_{xto0}frac{e^{-1/x^2}}x=lim_{xtopminfty}e^{-x^2}{x}=lim_{xtopminfty}frac{x}{e^{x^2}}=lim_{xtopminfty}frac{1}{2xe^{x^2}}$$
            Explaination
            $$lim_{xto0}f(x)=lim_{xtopminfty}f(1/x)$$
            It can be easily proved by using epsilon-delta language.

            I applied L'hospital in the last equal sign.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I'm sorry this may sound embarrassing, even though I'm a 3rd year math major. Can you elaborate more on your hint? How does $lim_{xto0}frac{e^{-1/x^2}}x$ becomes $lim_{xtoinfty}e^{-x^2}{x}$?
              $endgroup$
              – Jeffry Santosa
              Dec 7 '18 at 6:12












            • $begingroup$
              If you need more explanation, just ask.
              $endgroup$
              – Kemono Chen
              Dec 7 '18 at 6:17










            • $begingroup$
              Note that $lim_{xto 0} f(x) = lim_{x to infty} f(1/x)$ is not true in general. If the function is not continuous at zero the latter limit may still exist.
              $endgroup$
              – Pjotr5
              Dec 7 '18 at 8:09










            • $begingroup$
              @Pjotr5 Here $xtoinfty$ denotes $xtopminfty$.
              $endgroup$
              – Kemono Chen
              Dec 7 '18 at 8:10












            • $begingroup$
              @KemonoChen To be very precise it would be good to mention this, because that is not standard notation for real limits.
              $endgroup$
              – Pjotr5
              Dec 7 '18 at 8:19














            1












            1








            1





            $begingroup$

            Hint
            $$lim_{xto0}frac{e^{-1/x^2}}x=lim_{xtopminfty}e^{-x^2}{x}=lim_{xtopminfty}frac{x}{e^{x^2}}=lim_{xtopminfty}frac{1}{2xe^{x^2}}$$
            Explaination
            $$lim_{xto0}f(x)=lim_{xtopminfty}f(1/x)$$
            It can be easily proved by using epsilon-delta language.

            I applied L'hospital in the last equal sign.






            share|cite|improve this answer











            $endgroup$



            Hint
            $$lim_{xto0}frac{e^{-1/x^2}}x=lim_{xtopminfty}e^{-x^2}{x}=lim_{xtopminfty}frac{x}{e^{x^2}}=lim_{xtopminfty}frac{1}{2xe^{x^2}}$$
            Explaination
            $$lim_{xto0}f(x)=lim_{xtopminfty}f(1/x)$$
            It can be easily proved by using epsilon-delta language.

            I applied L'hospital in the last equal sign.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 7 '18 at 8:21

























            answered Dec 7 '18 at 6:04









            Kemono ChenKemono Chen

            3,0521743




            3,0521743












            • $begingroup$
              I'm sorry this may sound embarrassing, even though I'm a 3rd year math major. Can you elaborate more on your hint? How does $lim_{xto0}frac{e^{-1/x^2}}x$ becomes $lim_{xtoinfty}e^{-x^2}{x}$?
              $endgroup$
              – Jeffry Santosa
              Dec 7 '18 at 6:12












            • $begingroup$
              If you need more explanation, just ask.
              $endgroup$
              – Kemono Chen
              Dec 7 '18 at 6:17










            • $begingroup$
              Note that $lim_{xto 0} f(x) = lim_{x to infty} f(1/x)$ is not true in general. If the function is not continuous at zero the latter limit may still exist.
              $endgroup$
              – Pjotr5
              Dec 7 '18 at 8:09










            • $begingroup$
              @Pjotr5 Here $xtoinfty$ denotes $xtopminfty$.
              $endgroup$
              – Kemono Chen
              Dec 7 '18 at 8:10












            • $begingroup$
              @KemonoChen To be very precise it would be good to mention this, because that is not standard notation for real limits.
              $endgroup$
              – Pjotr5
              Dec 7 '18 at 8:19


















            • $begingroup$
              I'm sorry this may sound embarrassing, even though I'm a 3rd year math major. Can you elaborate more on your hint? How does $lim_{xto0}frac{e^{-1/x^2}}x$ becomes $lim_{xtoinfty}e^{-x^2}{x}$?
              $endgroup$
              – Jeffry Santosa
              Dec 7 '18 at 6:12












            • $begingroup$
              If you need more explanation, just ask.
              $endgroup$
              – Kemono Chen
              Dec 7 '18 at 6:17










            • $begingroup$
              Note that $lim_{xto 0} f(x) = lim_{x to infty} f(1/x)$ is not true in general. If the function is not continuous at zero the latter limit may still exist.
              $endgroup$
              – Pjotr5
              Dec 7 '18 at 8:09










            • $begingroup$
              @Pjotr5 Here $xtoinfty$ denotes $xtopminfty$.
              $endgroup$
              – Kemono Chen
              Dec 7 '18 at 8:10












            • $begingroup$
              @KemonoChen To be very precise it would be good to mention this, because that is not standard notation for real limits.
              $endgroup$
              – Pjotr5
              Dec 7 '18 at 8:19
















            $begingroup$
            I'm sorry this may sound embarrassing, even though I'm a 3rd year math major. Can you elaborate more on your hint? How does $lim_{xto0}frac{e^{-1/x^2}}x$ becomes $lim_{xtoinfty}e^{-x^2}{x}$?
            $endgroup$
            – Jeffry Santosa
            Dec 7 '18 at 6:12






            $begingroup$
            I'm sorry this may sound embarrassing, even though I'm a 3rd year math major. Can you elaborate more on your hint? How does $lim_{xto0}frac{e^{-1/x^2}}x$ becomes $lim_{xtoinfty}e^{-x^2}{x}$?
            $endgroup$
            – Jeffry Santosa
            Dec 7 '18 at 6:12














            $begingroup$
            If you need more explanation, just ask.
            $endgroup$
            – Kemono Chen
            Dec 7 '18 at 6:17




            $begingroup$
            If you need more explanation, just ask.
            $endgroup$
            – Kemono Chen
            Dec 7 '18 at 6:17












            $begingroup$
            Note that $lim_{xto 0} f(x) = lim_{x to infty} f(1/x)$ is not true in general. If the function is not continuous at zero the latter limit may still exist.
            $endgroup$
            – Pjotr5
            Dec 7 '18 at 8:09




            $begingroup$
            Note that $lim_{xto 0} f(x) = lim_{x to infty} f(1/x)$ is not true in general. If the function is not continuous at zero the latter limit may still exist.
            $endgroup$
            – Pjotr5
            Dec 7 '18 at 8:09












            $begingroup$
            @Pjotr5 Here $xtoinfty$ denotes $xtopminfty$.
            $endgroup$
            – Kemono Chen
            Dec 7 '18 at 8:10






            $begingroup$
            @Pjotr5 Here $xtoinfty$ denotes $xtopminfty$.
            $endgroup$
            – Kemono Chen
            Dec 7 '18 at 8:10














            $begingroup$
            @KemonoChen To be very precise it would be good to mention this, because that is not standard notation for real limits.
            $endgroup$
            – Pjotr5
            Dec 7 '18 at 8:19




            $begingroup$
            @KemonoChen To be very precise it would be good to mention this, because that is not standard notation for real limits.
            $endgroup$
            – Pjotr5
            Dec 7 '18 at 8:19











            1












            $begingroup$

            We have
            $$lim_{xrightarrow 0}frac{e^{-1/x^2}}{x}=lim_{trightarrow infty}frac{t}{e^{t^2}}=lim_{trightarrow infty}frac{1}{2te^{t^2}}=0 $$
            by $t=frac{1}{x}$ and L'Hospital's rules.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              We have
              $$lim_{xrightarrow 0}frac{e^{-1/x^2}}{x}=lim_{trightarrow infty}frac{t}{e^{t^2}}=lim_{trightarrow infty}frac{1}{2te^{t^2}}=0 $$
              by $t=frac{1}{x}$ and L'Hospital's rules.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                We have
                $$lim_{xrightarrow 0}frac{e^{-1/x^2}}{x}=lim_{trightarrow infty}frac{t}{e^{t^2}}=lim_{trightarrow infty}frac{1}{2te^{t^2}}=0 $$
                by $t=frac{1}{x}$ and L'Hospital's rules.






                share|cite|improve this answer









                $endgroup$



                We have
                $$lim_{xrightarrow 0}frac{e^{-1/x^2}}{x}=lim_{trightarrow infty}frac{t}{e^{t^2}}=lim_{trightarrow infty}frac{1}{2te^{t^2}}=0 $$
                by $t=frac{1}{x}$ and L'Hospital's rules.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 7 '18 at 6:07









                LauLau

                527315




                527315






























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