If $f$ is differentiable at $x = x_0$ then $f$ is continuous at $x = x_0$.

Question

Claim: if $f$ is differentiable at $x = x_0$ then $f$ is continuous at $x = x_0$.

Please, see if I made some mistake in the proof below. I mention some theorems in the proof:

The condition to $f(x)$ be continuous at $x=x_0$ is $\lim\limits_{x\to x_0} f(x)=f(x_0)$.

(1) If $f(x)$ is differentiable at $x-x_0$, then $f'(x)=\lim\limits_{x\to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}$ exists and the function is defined at $x=x_0$.

(2) Therefore, by the Limit Linearity Theorem, $\lim\limits_{x\to x_0} f(x)$ exists and we'll show it is equals $f(x_0)$.

(3) We'll do this by the Precise Limit Definion: given $ \epsilon>0, \exists\delta|0<|x-x_0|<\delta$, then $0<|f(x)-f(x_0)|<\epsilon$. As this limit exists by (2), we can make $f(x)$ as close to $f(x_0)$ as one wishes, therefore $\lim\limits_{x\to x_0} f(x)=f(x_0)$, what satisfies the condition for $f(x)$ be differentiable at $x=x_0$. The end.

Answer 1

$$\lim\limits_{x\to x_0}f(x)=\lim\limits_{x\to x_0}\left(f(x_0)+(x-x_0)\cdot\dfrac {f(x)-f(x_0)}{x-x_0}\right)=$$ $$\lim\limits_{x\to x_0}f(x_0)+\lim\limits_{x\to x_0}(x-x_0)\cdot\lim\limits_{x\to x_0}\dfrac {f(x)-f(x_0)}{x-x_0}=$$ $$f(x_0)+0\cdot f'(x_0)=f(x_0)$$

Answer 2

You surely mess up things in the proof. You say we can make $f(x)$ as close to $f(x_0)$ as one whishes. But this is exactly what you are trying to prove.

The more useful definition of differentiability is, that a function is differentiable if $f(x)=f(x_0) + a \cdot (x-x_0) + r(x)$ with $$\lim_{x\to x_0} \frac{r(x)}{x-x_0} = 0 $$ so $\lim_{x\to x_0} r(x)$ will surely be $0$. Hence \begin{align*} |f(x)-f(x_0)|&= | f(x_0) + a \cdot (x-x_0) + r(x) -f(x_0)|\\ &\leq |a| \cdot |x-x_0|+ |r(x)| \end{align*} Now you have a sum of 2 terms which go to zero as $x$ goes to $x_0$.

Answer 3

Here, a solution I've found at "Introduction to Analysis" byt Arthur Mattuck:

$$\lim_{x\to x_0} (f(x)-f(x_0)) = \lim_{x\to x_0} (\dfrac{f(x)-f(x_0)}{x-x_0})(x-x_0)\\ = \lim_{x\to x_0} (\dfrac{f(x)-f(x_0)}{x-x_0}) \cdot \lim_{x\to x_0} (x-x_0)\\ = f'(x)\cdot 0 \\ =0$$