$int _0^infty f(x) $ exists and $f(x)$ is differentiable then $lim _{x to infty} f '(x)$ exists. Counter...












7














Can anyone give me a counter example of the statement




If $int_0^infty f(x) $ exists and $f(x)$ is differentiable then $lim _{x to infty} f'(x)$ exists.




My attempt: I have thought one. First I draw $1/x^2$ in the first quadrant and $-1/x^2$ in the fourth quadrant. The area under the following curves are finite.



1) $1/x^2$



2) $-1/x^2$



3) $x= 1$.



Now I have drawn infinite number of $y = x+c $ at equal distances in that region. Then I joined those infinite lines by some smooth curve so that the curve remains differentiable. Now I think this function can be a counter example.



I am uploading one picture of my attempt. Can anyone please check it and if possible suggest me a better function.enter image description here










share|cite|improve this question





























    7














    Can anyone give me a counter example of the statement




    If $int_0^infty f(x) $ exists and $f(x)$ is differentiable then $lim _{x to infty} f'(x)$ exists.




    My attempt: I have thought one. First I draw $1/x^2$ in the first quadrant and $-1/x^2$ in the fourth quadrant. The area under the following curves are finite.



    1) $1/x^2$



    2) $-1/x^2$



    3) $x= 1$.



    Now I have drawn infinite number of $y = x+c $ at equal distances in that region. Then I joined those infinite lines by some smooth curve so that the curve remains differentiable. Now I think this function can be a counter example.



    I am uploading one picture of my attempt. Can anyone please check it and if possible suggest me a better function.enter image description here










    share|cite|improve this question



























      7












      7








      7


      3





      Can anyone give me a counter example of the statement




      If $int_0^infty f(x) $ exists and $f(x)$ is differentiable then $lim _{x to infty} f'(x)$ exists.




      My attempt: I have thought one. First I draw $1/x^2$ in the first quadrant and $-1/x^2$ in the fourth quadrant. The area under the following curves are finite.



      1) $1/x^2$



      2) $-1/x^2$



      3) $x= 1$.



      Now I have drawn infinite number of $y = x+c $ at equal distances in that region. Then I joined those infinite lines by some smooth curve so that the curve remains differentiable. Now I think this function can be a counter example.



      I am uploading one picture of my attempt. Can anyone please check it and if possible suggest me a better function.enter image description here










      share|cite|improve this question















      Can anyone give me a counter example of the statement




      If $int_0^infty f(x) $ exists and $f(x)$ is differentiable then $lim _{x to infty} f'(x)$ exists.




      My attempt: I have thought one. First I draw $1/x^2$ in the first quadrant and $-1/x^2$ in the fourth quadrant. The area under the following curves are finite.



      1) $1/x^2$



      2) $-1/x^2$



      3) $x= 1$.



      Now I have drawn infinite number of $y = x+c $ at equal distances in that region. Then I joined those infinite lines by some smooth curve so that the curve remains differentiable. Now I think this function can be a counter example.



      I am uploading one picture of my attempt. Can anyone please check it and if possible suggest me a better function.enter image description here







      limits analysis derivatives proof-verification continuity






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      edited Nov 22 at 5:07









      gt6989b

      32.8k22452




      32.8k22452










      asked Nov 22 at 4:57









      cmi

      1,000212




      1,000212






















          1 Answer
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          8














          Yep, this works perfectly! You can rigorize this sort of idea by defining some function like



          $$frac{sinleft(x^{10}right)}{x^2}$$



          (where the exponent of $10$ is simply to make sure our function oscillates fast enough).






          share|cite|improve this answer

















          • 1




            Actually simply $sin (x^2)$ suffices.
            – Szeto
            Nov 22 at 5:24










          • @Szeto I imagined lower exponents sufficed, but I didn't want to do out the calculations :-)
            – Carl Schildkraut
            Nov 22 at 5:47






          • 1




            Well, this function is in my memory because of the Fresnel’s integral, which is the first integral I see that converges while its integrand does not vanish. :)
            – Szeto
            Nov 22 at 5:49











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          1 Answer
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          active

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          active

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          active

          oldest

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          8














          Yep, this works perfectly! You can rigorize this sort of idea by defining some function like



          $$frac{sinleft(x^{10}right)}{x^2}$$



          (where the exponent of $10$ is simply to make sure our function oscillates fast enough).






          share|cite|improve this answer

















          • 1




            Actually simply $sin (x^2)$ suffices.
            – Szeto
            Nov 22 at 5:24










          • @Szeto I imagined lower exponents sufficed, but I didn't want to do out the calculations :-)
            – Carl Schildkraut
            Nov 22 at 5:47






          • 1




            Well, this function is in my memory because of the Fresnel’s integral, which is the first integral I see that converges while its integrand does not vanish. :)
            – Szeto
            Nov 22 at 5:49
















          8














          Yep, this works perfectly! You can rigorize this sort of idea by defining some function like



          $$frac{sinleft(x^{10}right)}{x^2}$$



          (where the exponent of $10$ is simply to make sure our function oscillates fast enough).






          share|cite|improve this answer

















          • 1




            Actually simply $sin (x^2)$ suffices.
            – Szeto
            Nov 22 at 5:24










          • @Szeto I imagined lower exponents sufficed, but I didn't want to do out the calculations :-)
            – Carl Schildkraut
            Nov 22 at 5:47






          • 1




            Well, this function is in my memory because of the Fresnel’s integral, which is the first integral I see that converges while its integrand does not vanish. :)
            – Szeto
            Nov 22 at 5:49














          8












          8








          8






          Yep, this works perfectly! You can rigorize this sort of idea by defining some function like



          $$frac{sinleft(x^{10}right)}{x^2}$$



          (where the exponent of $10$ is simply to make sure our function oscillates fast enough).






          share|cite|improve this answer












          Yep, this works perfectly! You can rigorize this sort of idea by defining some function like



          $$frac{sinleft(x^{10}right)}{x^2}$$



          (where the exponent of $10$ is simply to make sure our function oscillates fast enough).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 5:02









          Carl Schildkraut

          11.1k11441




          11.1k11441








          • 1




            Actually simply $sin (x^2)$ suffices.
            – Szeto
            Nov 22 at 5:24










          • @Szeto I imagined lower exponents sufficed, but I didn't want to do out the calculations :-)
            – Carl Schildkraut
            Nov 22 at 5:47






          • 1




            Well, this function is in my memory because of the Fresnel’s integral, which is the first integral I see that converges while its integrand does not vanish. :)
            – Szeto
            Nov 22 at 5:49














          • 1




            Actually simply $sin (x^2)$ suffices.
            – Szeto
            Nov 22 at 5:24










          • @Szeto I imagined lower exponents sufficed, but I didn't want to do out the calculations :-)
            – Carl Schildkraut
            Nov 22 at 5:47






          • 1




            Well, this function is in my memory because of the Fresnel’s integral, which is the first integral I see that converges while its integrand does not vanish. :)
            – Szeto
            Nov 22 at 5:49








          1




          1




          Actually simply $sin (x^2)$ suffices.
          – Szeto
          Nov 22 at 5:24




          Actually simply $sin (x^2)$ suffices.
          – Szeto
          Nov 22 at 5:24












          @Szeto I imagined lower exponents sufficed, but I didn't want to do out the calculations :-)
          – Carl Schildkraut
          Nov 22 at 5:47




          @Szeto I imagined lower exponents sufficed, but I didn't want to do out the calculations :-)
          – Carl Schildkraut
          Nov 22 at 5:47




          1




          1




          Well, this function is in my memory because of the Fresnel’s integral, which is the first integral I see that converges while its integrand does not vanish. :)
          – Szeto
          Nov 22 at 5:49




          Well, this function is in my memory because of the Fresnel’s integral, which is the first integral I see that converges while its integrand does not vanish. :)
          – Szeto
          Nov 22 at 5:49


















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