Find the fallacy in the resoning of showing $\lim\limits_{x\to0} \cos\frac{1}{x} =0$

Question

Question The function$$f(x)=\begin{cases} x^{2} \sin \frac{1}{x} & \text {for } x \neq 0 \\ 0 & \text {for } x=0 \end{cases}$$ is differentiable for any $x$. By Lagrange's Mean Value Theorem$$x^{2} \sin \frac{1}{x}=x\left(2 \xi \sin \frac{1}{\xi}-\cos \frac{1}{\xi}\right)\text{ where }0 <\xi <x$$Hence $$\cos \frac{1}{\xi}=2 \xi \sin \frac{1}{\xi}-x \sin \frac{1}{x} $$ As $x$ tends to zero, $\xi$ will also tend to zero. Passing to the limit, we obtain $$\lim\limits_{\xi \to 0} \cos (1 / \xi)=0$$ whereas it is known that $\lim \limits_{x \to 0} \cos \frac{1}{x}$ is non-existent.

Find the mistake in this reasoning. This proof seems logical but this is not valid. Why is this?

Answer 1

Your argument does not prove that $\lim_{\xi \to 0} \cos (\frac 1 {\xi})=0$ It only proves that $0$ is a limit point of $ \cos (\frac 1 {\xi})$. The reason is $\xi$ is not arbitrary. Though it tends to $0$ as $x \to 0$ it need not take all values near $0$.

Answer 2

Your line $$x^{2} \sin \frac{1}{x}=x\left(2 \xi \sin \frac{1}{\xi}-\cos \frac{1}{\xi}\right)\text{ where }0 <\xi <x$$ is mathematically illiterate, and I think is the source of your problem. It should read $$\text{There exists }\xi\in(0,x)\text{ such that }x^{2} \sin \frac{1}{x}=x\left(2 \xi \sin \frac{1}{\xi}-\cos \frac{1}{\xi}\right)$$ or alternatively $$x^{2} \sin \frac{1}{x}=x\left(2 \xi \sin \frac{1}{\xi}-\cos \frac{1}{\xi}\right)\text{ for some }0 <\xi <x$$ Your subsequent argument seems to assume that this holds for an arbitrary $\xi\in(0,x)$, which is not the case.

Answer 3

Note that $\xi$ is not a free value but a particular value between $0$ and $x$, dependent upon $x$, such that the original identity holds and such that the limit for $\cos \frac1{\xi}$ is equal to zero.

We should write

$$0< \xi(x)<x \implies \lim_{x\to 0} \xi(x)<0$$

and

$$\lim_{x\to 0}\cos \frac{1}{\xi(x)}=\lim_{x\to 0} \left(2 \xi(x) \sin \frac{1}{\xi(x)}-x \sin \frac{1}{x} \right)=0$$