If $\lim_{x\to+\infty}f'(x)>0$, does that mean that $\lim_{x\to+\infty}f(x)=+\infty$ Let's assume some function $f: \mathbb{R}\to\mathbb{R}$, which is differentiable in some region $(a, +\infty)$, where $a\in(0,+\infty)$. It makes sense (to me) that the following holds true:
$$\lim_{x\to+\infty}f'(x)\in(0, +\infty) \Rightarrow \lim_{x\to+\infty}f(x)=+\infty$$
However, I am not quite sure how I would go about proving it, if it even is true.
 A: If $\lim_{x\to\infty}f'(x)=\ell >0$, there exists $A\in\mathbf R$ such that $f'(x)>\dfrac\ell2$ for all $x>A$.
Then, for all such $x$s, apply the Mean value theorem between $A$ and $x$:
There exists some $ξ\in(A, x)$, such that
$$f'(ξ)=\frac{f(x)-f(A)}{x-A}>\frac{\ell}{2} \Rightarrow f(x)>\frac{\ell (x-A)}{2} + f(A)$$
Also
$$\lim_{x\to+\infty}\frac{\ell(x-A)}{2}=+\infty$$
And, since $f(A)\in\mathbb{R}$ $$\lim_{x\to+\infty}f(x)=+\infty$$
A: Sketch:
Assuming that $\lim_{x\rightarrow\infty}f'(x)$ exists and equals $L>0$, then there is some $M$ such that for all $x>M$, $|f'(x)-L|>\frac{L}{2}$.  Therefore, $f'(x)>\frac{L}{2}$ for $x>M$.
Observe that 
$$
f(x)=f(M)+\int_M^xf'(t)dt.
$$
For $x>M$, it follows that
$$
f(x)=f(M)+\int_M^xf'(t)dt\geq f(M)+\int_M^x\frac{L}{2}dt=f(M)+\frac{L}{2}(x-M).
$$
Since the RHS diverges to infinity, so does $f(x)$.
A: It is true.
First of all, call $L= \lim_{x \to \infty} f'(x)$. Since $L >0$, $f'$ is eventually positive, thus $f$ is eventually increasing.
Monotone functions always have a limit, thus there exists $a= \lim_{x \to \infty} f(x)$ ($a \in (0, \infty]$).
If $a \neq \infty$, you would have an horizontal asymptote (i.e. $L=0$), so necessarily $a= \infty$.
