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I am confused as to how to disprove the differentiability of a function. Let's, for example, consider the function $$f(x)=\begin{cases} x\cos\left(\frac{1}{x}\right), & \text{for }x\neq0,\\ 0 & \text{for }x=0. \end{cases}$$

One can easily see that $$\frac{f(x)-f(0)}{x-0}=\frac{x\cos\left(\frac{1}{x}\right)}{x}=\cos\left(\frac{1}{x}\right)$$ and thus it can't be differentiable at $0$ since the limit as $x\to 0$ of the above equation doesn't exist.

Question: What is necessary exactly? I've seen things like constructin two sequences $a_{n}=\frac{1}{2\pi n}$ and $b_{n}=\frac{1}{(2n+1)\pi}$ that converge to $0$ but $f'(a_n)=1$ and $f'(b_n)=-1$. Why isn't one sequence enough? I thought that a limit $$A:=\lim_{x\to a} f(x)$$ exists if and only if for every sequence $a_n$ that converges to $a$ we have $$\lim_{n\to\infty} f(a_n)=A.$$ So one sequence for the above example where $f(a_n)\neq 0$ should suffice in my eyes. Why are usually two sequences taken?

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To prove that $\lim_{x\to a}F(x)$ Exists, you will need to show that for any sequence $(a_n)$ that converges to $ a$ the sequence $(F(a_n))$ converges. But

if you want to prove that this limit $(\lim_{x\to a}F(x) ),$ DOES NOT EXIST, you could exhibit two sequences $(a_n)$ and $(b_n)$ which both goes to $ a$ and such that

the sequences of images $(F(a_n)) $ and $(F(b_n))$ have two different limit.

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