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Let $f:(0,\infty)\rightarrow\mathbb{R}$ be a differentiable function such that$\ \lim_{x \rightarrow \infty} f^{\ '}(x)=L>0$. Prove $f$ is unbounded.

My try:

$\lim_{x \rightarrow \infty} f^{\ '}(x)=L>0\Rightarrow$ Choosing $\epsilon=\frac{L}{2}$, we get $\exists x_0 \ \forall x>x_0:f^{\ '}(x)>\frac{L}{2}.$

By MVT: $f(x)>\frac{L}{2}x +f(0)$.

Is my proof correct, and can I conclude from it the result?

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Not quite; your last step assumes that $f'(x) > \frac{L}{2}$ for all $x > 0$. It is true that $f(x) > \frac{L}{2} (x-x_0) + f(x_0)$, though, and this is good enough.

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take $a$ to be large enough so that $f^{\prime}(a) \in B(L,\epsilon)$ for all $y \geq a$. Well, then $f(y) \geq f(a)+(L/2)(y-a)$, which is unbounded.

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