If I is an open interval, f is differentiable on I and a ∈ I, then there exists a sequence $a_n$ ∈ I such that $a_n → a$ and $f '(a_n) → f' (a).$ Hint: Use MVT

I know it seems simple but I just can't get it right.

$|a_n - a| \leq \epsilon$ then $ |f(a_n) - f(a)| \leq \epsilon $ also I get that the Mean Value Theorem states that for an interval $(a,b)$ if f is continuous on the interval then there exist $x_0 \in (a,b)$ st $f'(x_0)=\frac{f(b)-f(a)}{b-a}$

How I understand the problem is it that if the interval $(a,b)$ is $(a_n, a)$ then for any $x_1 \in (a_n,a)$ and $x_2 \in (a_n, a)$ then $f'(x_1)=\frac{f(a_n)-f(a)}{a_n-a}=f'(x_2)$ since interval $(a_n,a)$ is infinitely small as $n \rightarrow \infty$ then we kind of get $f'(a)=\frac{f(a_n)-f(a)}{a_n-a}=f'(a_n)$ but that feels just wrong. Can someone help me complet this properly?


We have for each $n$ some $|h_{n}|>0$ that $\left|\dfrac{f(a+h_{n})-f(a)}{h_{n}}-f'(a)\right|<\dfrac{1}{n}$, we may choose those $|h_{n}|$ such that $h_{n}\rightarrow 0$.

Write $\dfrac{f(a+h_{n})-f(a)}{h_{n}}=f'(a_{n})$ and we note that $a_{n}$ is in between $a+h_{n}$ and $a$, so $a_{n}\rightarrow a$ since $h_{n}\rightarrow 0$.


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