Is the function differentiable at $0$?

Question

Let $$f(x) = \begin{cases}\begin{align*}&\cos{\dfrac{1}{x}}, &x \neq0 \\ &0, &x=0. \end{align*}\end{cases}$$

Is the function $F(x) = \displaystyle \int_{0}^x f dx$ differentiable at $0$?

We can see that the function $f(x)$ is continuous everywhere except $x=0$. In order to show differentiability we will need to show the derivatives from the left and right are equal. So we need to show that $$\lim_{h \to 0^+} \dfrac{F(x+h)-F(x)}{h} = \lim_{h \to 0^-} \dfrac{F(x+h)-F(x)}{h}.$$ The derivative from the right is $$\lim_{h \to 0^+} \dfrac{F(x+h)-F(x)}{h} = \lim_{h \to 0^+} \dfrac{\displaystyle \int_{0}^x\cos{\dfrac{1}{x+h}}-\int_{0}^x\cos{\dfrac{1}{x}}}{h}$$ and the derivative from the left is $$\lim_{h \to 0^-} \dfrac{F(x+h)-F(x)}{h} = \lim_{h \to 0^-} \dfrac{\displaystyle \int_{0}^x\cos{\dfrac{1}{x+h}}-\int_{0}^x\cos{\dfrac{1}{x}}}{h}.$$ I am not sure how to proceed.

I think I am misreading something, but are your difference quotients for $f(x)$ or $F(x)$ (which is the antiderivative of $f(x)$)? — Chill2Macht, May 14 '16 at 02:11
very good question +1 Normally continuity of $f$ at $c$ guarantees that $F'(c) = f(c)$ and a jump discontinuity of $f$ guarantees non-differentiability of $F$. But here $f$ is oscillating at $0$. — Paramanand Singh, May 14 '16 at 05:49
I remember throwing the book against the wall after 4 hours working on this problem...........lol — Mathemagician1234, May 14 '16 at 05:52
@Mathemagician1234: I had done the problem long back from Spivak and hence remembered the old solution. You should not have thrown Spivak's book because it had a nice hint which had helped me come up with solution. :) — Paramanand Singh, May 14 '16 at 05:59
By the way your equations as you have written them are wrong. First you need to write your integrals with some other variable say $t$ because $x$ or $h$ are supposed to appear in the upper/lower limit of integral. Thus $$F'(x) = \lim_{h \to 0}\frac{F(x + h) - F(x)}{h} = \lim_{h \to 0}\frac{1}{h}\left(\int_{0}^{x + h}\cos(1/t),dt - \int_{0}^{x}\cos(1/t),dt\right)$$ — Paramanand Singh, May 14 '16 at 06:05
@Timkinsella: Problem is from Spivak page 281 (see Problem 10) in the chapter on FTC and luckily Spivak gave a hint also on which my answer is based. — Paramanand Singh, May 14 '16 at 06:10
@ParamanandSingh Thanks! I thought "this is an introductory calc book? surely theres a hint for this." — Tim kinsella, May 14 '16 at 06:19

Paramanand Singh · Answer 1 · 2016-07-29T16:37:54.107

The problem is slightly tricky because we don't know the anti-derivative of $\cos (1/x)$ but we do know that function $g(x) = x^{2}\sin(1/x), g(0) = 0$ is continuous and differentiable for all $x$ and $$g'(x) = 2x\sin (1/x) - \cos(1/x),g'(0) = 0$$ and hence $$g(h) = \int_{0}^{h}(2t\sin (1/t) - \cos (1/t))\,dt$$ or $$\int_{0}^{h}\cos(1/t)\,dt = 2\int_{0}^{h}t\sin(1/t)\,dt - g(h)$$ and hence \begin{align} F'(0) &= \lim_{h \to 0}\frac{1}{h}\int_{0}^{h}\cos(1/t)\,dt\notag\\ &= \lim_{h \to 0}\frac{2}{h}\int_{0}^{h}t\sin(1/t)\,dt - \frac{g(h)}{h}\notag\\ &= 2\lim_{h \to 0}\frac{1}{h}\int_{0}^{h}t\sin(1/t)\,dt\notag\\ \end{align} Now $t\sin(1/t)$ has a removable discontinuity at $t = 0$ and its limit is $0$ as $t \to 0$ hence by Fundamental Theorem of Calculus the limit $$\lim_{h \to 0}\frac{1}{h}\int_{0}^{h}t\sin(1/t)\,dt$$ above is $0$ and therefore $F'(0) = 0$.

Tim kinsella · Answer 2 · 2016-05-14T08:05:14.560

3

For $x\geq 0$, let $[x]$ denote the largest element of $\mathbb{Z}\pi+\pi/2$ not greater than $x$.

$$\left|\int_0^x \cos(1/t)dt\right|=\left|\int_{1/x}^\infty \frac{ \cos(t)}{t^2}dt\right|\leq \left|\sum_{n=0}^\infty\int_{[1/x]+n\pi}^{[1/x]+(n+1)\pi}\frac{ \cos(t)}{t^2}dt\right|\leq \left|\int_{[1/x]}^{[1/x]+\pi}\frac{ \cos(t)}{t^2}dt\right|\leq$$

$$\left| [1/x]^{-2}\int_{[1/x]}^{[1/x]+\pi}\cos(t)dt\right|= 2[1/x]^{-2}\leq 2(1/x-\pi)^{-2}\leq cx^{2}$$

for some $c>0$ and all small enough $x$.

So $|F(x)|$ goes to zero like $x^2$ as $x\rightarrow 0^+$. So it has a right derivative which is zero. By symmetry the left derivative is also zero.

edited May 14 '16 at 08:05

answered May 14 '16 at 05:48

Tim kinsella

5,903

You managed to estimate the integral without using some sort of anti-derivative (like I did in my answer) +1 – Paramanand Singh May 14 '16 at 05:53
Also very nice +1 – RRL May 14 '16 at 07:02
1

$Cx^{-2}$ should be $Cx^2$, right? – Hans Lundmark May 14 '16 at 07:33
yes, thanks. fixed – Tim kinsella May 14 '16 at 07:52

score 0 · Answer 3 · answered May 14 '16 at 02:18

0

What you have is the derivative of f, not the derivative of F which is what you want. The derivative of F at x= 0, from the right, is $\lim_{h\to 0^+}\frac{F(h)- F(0)}{h}= \lim_{h\to 0^+}\frac{\int_0^h f(t)dt}{h}$.

answered May 14 '16 at 02:18

user247327

18,710

shouldn't this be a comment? – Tim kinsella May 14 '16 at 02:21
What about from the left? – user19405892 May 14 '16 at 02:29

score 0 · Answer 4 · answered May 17 '20 at 16:12

Since $\cos(1/x)$ is an even function, it suffices to compute

$$\lim_{h\to0^+}{1\over h}\int_0^h\cos(1/x)\,dx=\lim_{k\to\infty}k\int_k^\infty{\cos x\over x^2}\,dx=\lim_{k\to\infty}\left(-{\sin k\over k^2}-k\int_k^\infty{\sin x\over x^3}\,dx \right)$$

Now $\lim_{k\to\infty}(\sin k/k^2)=0$ and

$$\left|k\int_k^\infty{\sin x\over x^3}\,dx \right|\le k\int_k^\infty{dx\over x^3}={k\over2k^2}={1\over2k}\to0$$

It follows that $F'(0)=0$.

Is the function differentiable at $0$?

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