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