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Given a differentiable function $f: (0,\infty) \rightarrow \mathbb R $ and $c>0$ such that $f'(x)>c$ for every $x$.

Prove: $\lim _{ x\rightarrow\infty }{ f(x) }=\infty$

Using the MVT, I got to $f(x)>c(x-x_0)+f(x_0)$, and I think I should proceed using the definition of limit, but I got stuck.

Any help appreciated.

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    You've shown $f(x) > A + cx$ for some constant $A.$ Aren't you essentially done? – 2017-02-17
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    @zhw. Is it ok to say that $cx_0$ is constant? – 2017-02-17
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    Sure. Just take $x_0 = 1$ for example. – 2017-02-17
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    @zhw. Great thanks ! – 2017-02-17

1 Answers 1

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For $x>1$, we have that $f(x)-f(1)=\int_1^xf'(t)\,dt>c(x-1)$. Hence,

$$f(x)>c(x-1)+f(1)$$

And it's easy to see that the RHS is unbounded as $x\to\infty$.

EDIT: As @zhw. noted, the solution above does not work if $f'$ is not Riemann integrable. Nonetheless, we may use the general idea to craft a solution that does not involve integration.

Consider $g(x)=c(x-1)+f(1)$. Then $h(x)=f(x)-g(x)$ is such that $h(1)=0$ and $h'(x)>0$ for all $x>1$. It follows that $h(x)>0$ for all $x>1$, that is,

$$f(x)>c(x-1)+f(1)$$

for all $x>1$. The conclusion follows.

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    We don't know $f'$ is Riemann integrable. – 2017-02-17
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    Good observation. I have fixed it. – 2017-02-17
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    @Fimpellizieri Got you. Does choosing 1 require any further explanation ? Or do I need to say anything more after ? – 2017-02-17
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    Nope, you can choose any $t\in (0,+\infty)$. Of course, then $g(x)$ becomes $c(x-t)+f(t)$. – 2017-02-17
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    @Fimpellizieri Great, thanks ! – 2017-02-17