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I want to show that:

If $\lim\limits_{x \to \infty} f'(x)>0$ then $\lim\limits_{x \to \infty} f(x)= \infty$

My attempt: Since $\lim\limits_{x \to \infty} f'(x)>0$ there is some $\delta >0$ such that $\forall x>y\geq \delta \implies f(x)>f(y)$ i.e $f$ is increasing on $[\delta,\infty)$.

How to proceed?

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When $\lim_{x\to \infty} f'(x)>0$ then there is a $x_0$ such that $f'(x)>\varepsilon$ for all $x>x_0$ and some $\varepsilon>0$ with the mean value theorem we have $$f(x)-f(x_0)=f'(\xi) (x-x_0)$$ $x-x_0$ is not bounded and $f'(\xi)$ is strictly bigger zero, hence the righthand side will go to infinity when $x$ goes to infinity.

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Intuitively, if $\lim _{x\to \infty }f'(x)>0$, then for all large enough $x$ holds that $f'(x)>c>0$ for some constant $c$. That means that the slope of $f$ is at least $c$, and thus that the functions grows at least as fast as $cx$ does. Since $\lim _{x\to \infty }cx=\infty $, the result follows.

Now, take this intuitive proof and turn it into an actual proof by making all the steps precise. You'll want to use the mean value theorem to get a precise relation between values of $f$ and values of $f'$.

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