$f$ continuous, $a_n\neq2,\lim_{n \to \infty}a_n=2, b_n=\frac{f(a_n)-f(2)}{a_n-2}, b_n$ converges $\Rightarrow f$ is differentiable at $x=2$

Question

So I've been struggling with this prove/disprove question:

Let $f:\mathbb R\rightarrow \mathbb R$ be a continuous function and let $a_n\neq2, b_n $ be two sequences such that $\lim_{n \to \infty}a_n=2$ and $b_n=\frac{f(a_n)-f(2)}{a_n-2}.$ Prove or disprove that if $b_n$ converges, then $f$ is differentiable at $x=2$.

Since f is continuous, I realized that $\lim_{n \to \infty} f(a_n)=f(2)$, but that is essentially it. I can't seem to figure out how to deal with the "$\frac{0}{0}$" limit. It did occur to me, though, that this limit looks like a derivative, however I have no idea how to deal with it when it comes to sequences (Heine doesn't seem to help me much here).

Thank you very much and have a beautiful day.

Answer 1

The statement is false.

For example if you let $f(x) = |x - 2|$ and $a_n = 2 + \frac{1}{n}$ for each $n\in \mathbb{N}$, then $$\lim_{n\to\infty} a_n = 2$$ and $$\lim_{n\to\infty} b_n = \lim_{n\to\infty} \frac{f(a_n) - f(2)}{a_n - 2} = \lim_{n\to\infty} 1 = 1 $$ So $b_n$ converges, but $f$ is not differentiable at $2$.