If $f : (a,b) \to \mathbb{R}$ is differentiable, then $f$ is continuous: prove using $\epsilon$-$\delta$ Let $f : (a,b) \rightarrow \mathbb{R}$ be differentiable on $(a,b)$.
Prove that $f$ is continuous by showing that
$\forall x_0 \in (a,b) \qquad \forall \epsilon>0 \qquad \exists \delta >0  $ s.t.$ \; \;(x\in (a,b), |x-x_0| < \delta) \implies |f(x)-f(x_0)|<\epsilon$
This is a 6 mark question on a practice Analysis 2 module.
Frustratingly there are no solutions and I'm not sure where to start.
Any help is greatly appreciated.
 A: Observe that for all $x \neq x_0$, $x, x_0 \in (a,b)$, we have
$$
\begin{align}
|f(x) - f(x_0)| &= |f(x) - f(x_0)| - |(x - x_0)f'(x_0)| + |(x-x_0)f'(x_0)|\\
&\leq |f(x) - f(x_0) - (x-x_0)f'(x_0)| + |(x-x_0)f'(x_0)|\\
&= |x - x_0| \frac{|f(x) - f(x_0) - (x-x_0)f'(x_0)|}{|x-x_0|} + |(x-x_0)| |f'(x_0)|
\end{align}
$$
Now, for every $\epsilon > 0$ there exists $\delta > 0$ such that
$$
\left| \frac{f(x) - f(x_0) - (x-x_0)f'(x_0)}{x-x_0} \right| < \epsilon
$$
for all $|x - x_0| < \delta$. This is just a restatement of the existence of the derivative in terms of $\epsilon$'s and $\delta$'s.
So, for this $\epsilon > 0$, we have that
$$
|f(x) - f(x_0)| < \delta \epsilon + \delta |f'(x_0)|.
$$
Therefore, choosing $$\delta < \frac{\epsilon}{\epsilon + |f'(x_0)|},$$
we see that $|f(x) - f(x_0)| < \epsilon$ whenever $|x - x_0| < \delta$.
A: Fix $x_0 \in (a,b)$. $f$ is differentiable at $x_0$ means the limit $\lim_{x \rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0}=L$ exists.
By definition of limit, for any $\varepsilon>0$, there exists some $\delta>0$ such that $|\frac{f(x)-f(x_0)}{x-x_0}-L|<\varepsilon$ as $|x-x_0|<\delta$.
Can you rearrange the above inequality and use triangle inequality so that the above inequality becomes $|f(x)-f(x_0)|<M|x-x_0|<M\delta$ for some constant $M$?
Then the proof is done. (How?)
