Prove that $\lim_{n\to \infty} {f(x_n)-f(x_0)\over x_n - x_0}= f´(x_0)$

Question

Problem: Prove that if $f$ is continuous at $x_0$ and$$\lim_{n\to \infty} {f(x_n)-f(x_0)\over x_n - x_0}$$ exist for any sequence ${x_n} \to x_0$ and $x_n\neq x_0$ $\forall n\in \mathbb N$, then $f´(x_0)$ exists and $$\lim_{n\to \infty} {f(x_n)-f(x_0)\over x_n - x_0}= f´(x_0).$$

I started the proof like this: Let $\epsilon\gt 0$. Then there exists $N$ such that $\forall n\gt N$, $$\left|{f(x_n)-f(x_0)\over x_n - x_0}-L\right|\lt\epsilon.$$

By hypothesis $f$ is continuous at $x_0$, so there exist $\delta\gt 0$ such that $$\left|x-x_0\right|\lt\delta\Rightarrow \left|f(x)-f(x_0)\right|\lt\epsilon.$$

The elements in the sequence ${x_n}$ are in the neighborhood of radius $\delta$ for some $n\gt N$ but this was an arbitrary sequence so I don´t know if this implies that $f´(x_0)$ exist.

I would really appreciate your help with this problem thank you.

Answer 1

Hint: Assume that the function is not differentiable at $x_{0}$. Then check various ways in which a derivative fails to exist and correspondingly find a sequence $\{x_{n}\}$ for which limit in question does not exist. This contradiction solves the problem.

Answer 2

Suppose $f$ is not differentiable at $a := x_0$. Let $(x_n) \subset \mathbb{R} \setminus \{a\}$ be a sequence converging to $a$. If

$$\lim_{n \rightarrow \infty} \frac{f(x_n) - f(a)}{x_n - a}$$

does not exist we are already done. So suppose the limit does exist and equals $L \in \mathbb{R}$. Since $f$ is not differentiable at $a$ we can find a sequence $(y_n) \subset \mathbb{R} \setminus \{a\}$ converging to $a$ such that

$$\lim_{n \rightarrow \infty} \frac{f(y_n) - f(a)}{y_n - a} \not \rightarrow L$$

(else any sequence $(a_n) \subset \mathbb{R} \setminus \{a\}$ converging to $a$ would satisfy

$$\lim_{n \rightarrow \infty} \frac{f(a_n) - f(a)}{a_n - a} \rightarrow L$$

and we would have that $f$ is differentiable at $a$ with $f'(a) = L$. Please note carefully here that I am not assuming the required result, as the required result does not say all the limits for different sequences $(x_n)$ are the same, which is the contradiction we have arrived at here).

We can now define a new sequence $(z_n)$ by $z_{2n}:= x_n$ and $z_{2n+1}:= y_n$. Note that

$$ \frac{f(z_n) - f(a)}{z_n - a} $$

cannot converge (as it has a subsequence tending to $L$ and a subsequence which does not tend to L).

Answer 3

Definition

If $S$ is a subset of $\Bbb{R}$ and if $f:S\rightarrow\Bbb{R}$ is a function then for $x_0\in\overline{S\setminus\{x_0\}}$ we say that $f$ is derivable at $x_0$ if and only if exist the limit of the function $$ f'(x):=\frac{f(x)-f(x_0)}{x-x_0} $$ as $x$ approaches at $x_0$.

The set $\Bbb{R}$ equipped with the usual order topology is metrizable so that $\Bbb{R}$ is first countalbe and hausdorff separable.

Lemma

If $X$ is first countable and if $f:X\rightarrow Y$ is a function then $y_0$ is the limit of $f$ as $x$ approaches at $x_0$ if and only if for any sequence $(x_n)_{n\in\Bbb{N}}$ converging to $x_0$ it happens that $\big(f(x_n)\big)_{n\in\Bbb{N}}$ converges to $y_0$.

Proof. See here.

Lemma

If $Y$ is hausdorff separable and if $f:X\rightarrow Y$ is a function then for any limit point $x_0$ the limit of the function $f$ for $x$ approaches at $x_0$ is unique whether it exist.

Proof. It is sufficient to remember that any net with values in a hausdorff space can converges at most one point.

Theorem

If $S$ is a subset of $\Bbb{R}$ and if $f:S\rightarrow\Bbb{R}$ is a function and if for any sequence $(x_n)_{n\in\Bbb{N}}$ converging to $x_0$ the sequence derived sequence $f'(x_n)$ converges to the same $y_0$ then $f'(x_0)=y_0$.

Proof. It is an immediate consequence of the two previous lemmas.