# show that $f$ is differentiable for $x=0$

We have $f:\;]-1,1[\rightarrow\mathbb{R}$. $f$ is a continuous function on $]-1,1[$ and differentiable on $]-1,0[\cup]0,1[$. We suppose that $\lim_{x\rightarrow{0^{+}}}\frac{df(x)}{dx}=-\infty$. I need to show that $f$ is not differentiable for $x=0$.

The problem didn't seem to be solved and here are my thought about the question and the definition I have of a differentiable function.

A function is differentiable if $\forall\;x_{0}\in\;D$ we have $\lim_{x\rightarrow{x_{0}}}\frac{f(x)-f(x_{0})}{x-x_{0}}$ exists. In this case if the limit exists $\forall\;x_{0}\in\;]-1,1[$, then $f$ is differentiable. Because $f$ is differentiable, but not at zero, I was trying to consider that $f$ is differentiable at zero to get a contradiction. But I'm not sure where I can begin.

• If $f$ is differentiable, then the $\lim_{x\rightarrow 0^-}\frac{df}{dx}=-\infty$ (Otherwise the two sides limit are not agree). But if that is the case, the limit is unbounded, which imply the limit is not exist.
– Lion
Aug 6, 2015 at 0:30
• Apply the mean value theorem on the differential quotient on $]0,1[$ and show it converges to $-\infty$ using the assumption about the derivative. Aug 6, 2015 at 0:31
• I'm don't understand why the limit wouldn't exists. It would just mean that the limit is $-\infty$. I think the goal is to prove that there's no $\delta>0$ such as the limit converge to $-\infty$. Aug 6, 2015 at 0:43
• if the limit of the derivative at 0 is not finite this means that $f$ isnt differentiable right? Aug 6, 2015 at 0:45
• Yes, but this is what I'm trying to prove, with the given informations. Aug 6, 2015 at 0:48

Let $x\in ]0, 1[$. Then, $f$ is continuous on $[0, x]$ and differentiable on $]0, x[$. Thus, the mean value applies and there is some $c_x \in ]0, x[$ with $$\frac{f(x) - f(0)}{x - 0} = f'(c_x) \to -\infty.$$ Hence, the differential quotient has no finite limit and $f$ is not differentiable at $0$.
You can also apply directly the theorem of De l'Hospital to the limit $$\lim_{x\to 0}\frac{f(x)-f(0)}{x}.$$