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If $f(x)=\int_{0}^{x}\cos\frac{1}{t}\ {dt}$, then calculate $f'(0) $.

I found the answer there and understood its solution. But when I solve it myself, I first tried to solve it directly:

$$f'(0)=\frac{\int_{0}^{x}\cos\frac{1}{x}dx}{x},$$ And use the L'Hospital, I get $$f'(0)=\frac{\cos\frac{1}{x}}{1}$$ and get the conclusion that $f'(0)$ does not exist!

I do not know where is wrong, Could you point it out? Thank you!

Calvin Khor
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fractal
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  • L'Hopital is inconclusive when the resulting limit doesn't exist. – Brian Moehring Jul 02 '20 at 01:45
  • Thank you! Got it.@BrianMoehring – fractal Jul 02 '20 at 01:48
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    The Hospital is overrated. See https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule . The condition that $\lim f'(x)/g'(x)$ exists is a sufficient condition for $\lim f(x)/g(x)$ to exist in the indeterminate case, but as you see from the article it is never asserted it's a necessary condition. – Angina Seng Jul 02 '20 at 01:51
  • You should use Lebnitz rule – UmbQbify Jul 02 '20 at 02:01
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    @user675453 And how on earth does Leibniz's rule apply here? – Mark Viola Jul 02 '20 at 02:07
  • @Mark Viola, oh right... But it does give an expression for $f'(x)$ , which is not defined at 0. So is that the answer? – UmbQbify Jul 02 '20 at 02:15
  • No it isn't the answer. For $x\ne0$, $f'(x)=\cos(1/x)$. And hence, $\lim_{x\to 0}f'(x)$ fails to exist. BUT, $f'(0)=0$. So, $f'(x)$ is not continuous at $x=0$. – Mark Viola Jul 02 '20 at 02:28
  • @AsdrubalBeltran: your answer is not complete and does not do anything to clear out the problem. You may need to review the statement of the fundamental theorem of calculus. – Mittens Jul 02 '20 at 03:54

1 Answers1

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Let $f(x)$ be given by the integral

$$f(x)=\int_0^x \cos(1/t)\,dt$$

For $x\ne 0$, we see that $f'(x)=\cos(1/x)$. Hence, $\lim_{x\to0}f'(x)$ fails to exist.

But the fact that $\lim_{x\to0}f'(x)$ does not exist, does not imply that $f'(0)$ does not exist. In fact, we have

$$\begin{align} f'(0)&=\lim_{x\to 0}\frac{f(x)-f(0)}{x}\\\\ &=\lim_{x\to0}\frac{\int_0^x \cos(1/t)\,dt}x\\\\ &=\lim_{x\to 0}\left(-x\sin(1/x)+\frac1x\int_0^x 2t\sin(1/t)\,dt\right)\\\\ &=\lim_{x\to 0}\left(-x\sin(1/x)+O(x)\right)\\\\ &=0 \end{align}$$

Hence, we see that $f(x)$ is differentiable for all $x$, but its derivative, $f'(x)$, is discontinuous at $0$.

The conditions to apply L'Hospital's Rule apply. We have $f(x)=\int_0^x \cos(1/t)\,dt$ and $g(x)=x$ are differentiable on open interval's containing $0$. But the limit

$$\lim_{x\to 0}\frac{f'(x)}{g'(x)}=\lim_{x\to0}\frac{\cos(1/x)}{1}$$

does not exist. Therefore, L'Hospital's Rule is a useless tool to determine $f'(0)$.

And that is all that is going on here.

Mark Viola
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  • @OliverDiaz Yes, I integrated by parts with $u=t^2$ and $v=-\sin(1/t)$. – Mark Viola Jul 02 '20 at 03:25
  • @red_rose Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola Jul 02 '20 at 03:38
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    I think one should consider $\lim_{\varepsilon\rightarrow0}\int^x_\varepsilon\cos(t^{-1}),dt$ before integration by parts as $\sin(t^{-1})$ is not of finite variation. After this (although I am still checking) your argument should carry over. $$ \begin{align} \int^x_\varepsilon\cos(t^{-1}),dt&=-\int^x_\varepsilon t^2 d\sin(t^{-1}) \ &=-x^2\sin(x^{-1})+\varepsilon^2\sin(\varepsilon^{-1})+2\int^x_\varepsilon t\sin(t^{-1}),dt\ &\xrightarrow{\varepsilon\rightarrow0} -x^2\sin(x^{-1})+ 2\int^x_0 t\sin(t^{-1}),dt\ \end{align} $$ – Mittens Jul 02 '20 at 03:49
  • Just one comment on your explanation of L'Hosital rule. In the present case, one condition of L'Hosital rule fails, that is the existence of the limit $\frac{f'(x)}{g'(x)}$. Here is a version to clarify matters https://math.stackexchange.com/a/3727978/121671. That is the condition that an old professor of mine called the clairvoyant condition. – Mittens Jul 02 '20 at 05:09
  • Thank you for your correction! About if the FOC's application, could I think as followings? As $\cos\frac{1}{x}$ is continuous over $(0,x]$ for every x($x\neq 0$), and whatever integrand when $x=0$(as it is a single point, so don't affect the definite integral), so $\cos\frac{1}{x}$ is definitely integrable over $[0,x]$, and when over $[\delta,x]$, the $\cos\frac{1}{x}$ is continuous so it has antiderivative, so for any $x\neq 0$, there is an antiderivative. So using the part two of the FOC, the $f'(x)=\cos\frac{1}{x}$ when $x\neq 0$. – fractal Jul 02 '20 at 05:19
  • @red-rose Yes, you have it. Well done. – Mark Viola Jul 02 '20 at 13:58