if the $\lim_{x \to x_0} f(x) = \infty$ then $\lim_{x \to x_0} f'(x) = \infty$

Question

Let $x_0 ∈\Bbb R$, and let $f$ be a function defined on a neighborhood of $x_0$. Assume that $f$ is differentiable in a neighborhood of $x_0$, except maybe at $x_0$ itself, and assume that $\lim_{x\to x_0^-} f(x) = +∞$. Then $\lim_{x\to x_0^-}f'(x) = +∞$?

so ive been trying to come up with a counter example for hours and i cant tell if its true or not anymore...

Answer 1

Take $x_0 = 0$, and consider $f$ defined by $f(x) = \frac{1}{x^2}+\sin\frac{1}{x^2}$ (for $x\neq 0$) and $f(0)=0$. It fits the bill, but the derivative has no limit at $0^-$: $$ f'(x) = -\frac{2}{x^3}\left(1+\cos\frac{1}{x^2}\right), \qquad x\neq 0 $$