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I noticed that when $f'(x)$ tends to $+\infty$ as $x$ tends to $+\infty$, then $f(x)$ must tend to $+\infty$ as $x$ tends to $+\infty$ as well. I'm stuck at the proof though.

If you implement the mean value theorem for $f(x)$ in $(x,x+1)$ you'll get:

$f(x+1) - f(x)= f'(b)$ where $b>x$

now if u take limits for $x\rightarrow+\infty$, b must tend to $+\infty$ as well. So, $\lim_{x\to\infty}$ $f(x+1) - f(x)$ = $+\infty$, as $x$ tends to $+\infty$

Also, $f(x)$ must be increasing as $x$ tends to $+\infty$. Can somebody continue this or show me another way of proving this?

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marked as duplicate by 23rd, Rick Decker, Stefan Hansen, Lord_Farin, Dan Rust Jun 12 '13 at 13:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Since $f'(x)\to+\infty$ when $x\to+\infty$, there exists $x_0$ such that $f'(x)\geqslant1$ for every $x\geqslant x_0$. Thus, $f(x)\geqslant x-x_0+f(x_0)$ for every $x\geqslant x_0$. Since $x-x_0+f(x_0)\to+\infty$ when $x\to+\infty$, this implies that $f(x)\to+\infty$ when $x\to+\infty$.

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did u just think of that? ._. – Plom Jun 12 '13 at 12:34
@Plom This comes from the interpretation of $f'$ as the "gradient" of the graph: The graph gets steeper, so clearly it lies above a line with gradient 1 when $x$ is large enough. – user71815 Jun 12 '13 at 12:50
Plom: Are you able to expand on the "Thus" step? – Did Jun 12 '13 at 13:09
@Did What justifies the inequality following the word "Thus,"? – Jack M Jun 12 '13 at 13:15
@JackM Surely you can answer to that... :-) – Did Jun 12 '13 at 13:49

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