Property of the derivative of a function A function f has derivative for all $x\in \mathbb R$ and the limits of $f$ at $+\infty $, $-\infty$ are equal to  $+\infty$ . Is it true that $\lim_{x\to a} \frac {1}{f'(x)} = + \infty $ or $-\infty$  for some $a\in\mathbb R$ ?
Of course function $f' $ has roots , according to Fermat's theorem( $f$ has a total infimum) but how I could find an example to prove that the statement is W(wrong), if it really is wrong? 
Thank you in advance!
Babis
 A: Your statement is false. In fact you can take $f$ to be constant in some interval and let $f$ be decreasing before that interval and increasing after that interval. Thus let $f(x) =(x+1)^{2},x<-1,f(x)=0,|x|\leq 1,f(x)=(x-1)^{2},x>1$. Then we can see that $f$ is differentiable everywhere, but there is no point where $1/f'(x) \to \pm\infty$. 
A: Excuse me for(my poor english and ) comig again, I'm new here and I probably I do something in a wrong  way.There was a typo in my first messange and the problem exists yet.
My original question is to find if the function   $\frac {1}{f'(x)}$ has a vertical asymptote.
The function $f' $ has the Darboux property  and I'm trying to prove that the statement is false.But is it really fasle?
So , I'm trying to prove that  there is a function(or to construct  a function) , such that for all points  $a$ which are roots of $f'$ , both limits
$\lim_{x\to a^{+}}\frac {1}{f'(x)}$ ,$\lim_{x\to a^{-}}\frac {1}{f'(x)}$ 
are not equal to $+\infty $ or $ -\infty$  or these limits are not defined.
