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How can I prove that if $f$ is differential on $(0, \infty)$ and $f'$ is strictly growing on $(0,\infty)$ and, starting at some point $x_0 \in (0,\infty)$, $f'>0$, than $\lim_{x\to \infty} f(x) = \infty$?

It makes common sense yet I don't know how to approach this.

5xum
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1 Answers1

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If $f'(x_0) = M > 0$ and $f'$ is growing then $$ f'(x) \ge M \text{ for } x \ge x_0 $$ and therefore, using the mean-value theorem, $$ f(x) \ge f(x_0) + (x-x_0)M \text{ for } x \ge x_0 \, . $$

You could also argue that $f$ is convex and therefore its graph lies above the tangent line at $x=x_0$, which leads to the same result.

Martin R
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  • How does the fact that $f(x) \geq f(x_0) +(x-x_0)M$ imply that $lim_{x\to\infty}f(x)=\infty$? It would imply that if it was true for every $M>0$, but in this case it can't be said for every $M>0$, because there could be a situation where $f'$ has a limit somewhere so that can't be said. Am I missing something? –  Jan 23 '17 at 08:38
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    @S.Peter: The right-hand side $f(x_0) + (x-x_0)M$ has limit $+\infty$, and the left-hand side $f(x)$ is $\ge$ the right-hand side. – I did not say that this holds for every $M > 0$. I have chosen a fixed $M = f(x_0) > 0$ which exists according to your question. – Martin R Jan 23 '17 at 08:41