Let $f$ be defined in a deleted neighborhood of $U$ of $x_0 = 0$. $f$ is differentiable and has a vertical asymptote at $x = 0$ then at least one of the limits $\lim\limits_{x\rightarrow0^+} f'(x)$ and $\lim\limits_{x\rightarrow0^-} f'(x)$ exists, when they tends to infinity.
I think this statement is correct. if the limits were of $f(x)$ instead of $f'(x)$ then it was immediate by definition of an asymptote. my direction: I want to prove in contradiction that f converges and hopefully something will go wrong there. I'd like to get some help with that or another direction is good as well
Thanks in advance!
No, it's not true. For example, you can define a function which is differentiable for $x\neq 0$, increasing and tending to $+\infty$ as $x\to 0^+$ and decreasing to $-\infty$ as $x\to 0^-$, such that $f(1/n)=n$ and $f'(1/n)=0$ for every integer $n$. To do this, for each interval $(1/(n+1),1/n)$ you can choose a cubic equation with the required values and derivatives at $x=1/(n+1)$ and $x=1/n$, then just define $f(x)$ for $x\in(1/(n+1),1/n)$ to be the value of that cubic evaluated at $x$.