Is this an application of the mean value theorem? I have this problem and I was thinking to use the mean value theorem, but in the hypothesis i don't have the conditions (explicitly at least). Any ideas of how to start?

Suppose that $\lim_{x\to a} f(x) = 0$, and that exist $\lim_{x\to a} \frac{1}{f'(x)}$. Show that exist an open interval with center in $a$ such that in that interval but without the $a$ $f(x)≠0$.

 A: Since the limit of $1/f'$ exists the derivative $f'$ is non-zero in some deleted neighborhood of $a$ and therefore by intermediate value property it maintains a constant sign on each side of $a$ in this deleted neighborhood. Thus $f$ is strictly monotone on each side of $a$ in the deleted neighborhood. And since $f(a) =0$ it follows that $f$ is non-zero in the deleted neighborhood of $a$. 
A: Here's a proof idea which uses the MVT, or rather just its special case known as Rolle's Theorem.
Assume that for every $\epsilon >0$, there is $x\in (a, a+\epsilon)$ with $f(x)=0$.
Then for every $\epsilon$, there are actually infinitely many such $x$ (why?). Choose two of them, $x_{1,\epsilon}$ and $x_{2,\epsilon}$. By Rolle's Theorem (fill in details), $f'(x) = 0$ somewhere in the interval $(x_{1,\epsilon}, x_{2,\epsilon})$ which is contained in $(a, a+\epsilon)$. This holding for every $\epsilon$ contradicts the existence of $\lim_{x\to a}\frac{1}{f'(x)}$ (why?).
So there is some $\epsilon>0$ such that $f$ has no zero between $a$ and $a+\epsilon$. Completely analogously, find an interval to the left of $a$ without zeros of $f$.
