# Does one need two sequences to disprove differentiability?

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?

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.