Find the fallacy in the resoning of showing $\lim\limits_{x\to0} \cos\frac{1}{x} =0$ 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?
 A: 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 $0$.
A: 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.
A: 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
$$\lim_{x\to 0}\cos \frac{1}{\xi(x)}=\lim_{x\to 0} \left(2 \xi(x) \sin \frac{1}{\xi(x)}-x \sin \frac{1}{x} \right)=0$$
