# If $f : (a,b) \to \mathbb{R}$ is differentiable, then $f$ is continuous: prove using $\epsilon$-$\delta$ [duplicate]

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.

## marked as duplicate by Andrés E. Caicedo, Robert Soupe, Xander Henderson, Eckhard, user416281 Jun 19 '18 at 18:53

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 $|x - x_0| < \delta$.

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?)

• So all I had to do was plug a definition into a definition and rearrange? – Ben Crossley Jun 19 '18 at 11:55
• You can say so. – Jerry Jun 19 '18 at 11:56