Knowing the limit of $f'(x)$ find the limit of $f(x)$ We have that $f$ is differentiable on $(a, +\infty)$ with $a>0$. 
I want to show that if $\displaystyle{\lim_{x\rightarrow +\infty}f'(x)=\ell}$, then there are the following cases: 


*

*If $\ell>0$ then $$\lim_{x\rightarrow +\infty}f(x)=+\infty\ \text{ and } \ \lim_{x\rightarrow +\infty}\frac{f(x)}{x}=\ell$$ 

*If $\ell<0$ then $$\lim_{x\rightarrow +\infty}f(x)=-\infty\ \text{ and } \ \lim_{x\rightarrow +\infty}\frac{f(x)}{x}=\ell$$ 

*If $\ell=0$ then $$\lim_{x\rightarrow +\infty}f(x)=? \ \text{ and } \ \lim_{x\rightarrow +\infty}\frac{f(x)}{x}=\ell$$
$$$$ 
Do we use the fact that $$f(x+1)-f(x)=\int_x^{x+1}f'(u)\,du$$ to find the limits of the function $f$ ? 
 A: Let $ \varepsilon>0 : $
Since $ \lim\limits_{x\to +\infty}{f'\left(x\right)}=\ell $, there is some $ A_{1}\geq a $ such that $ \left(\forall x\geq A_{1}\right),\ \left|f'\left(x\right)-\ell\right|<\frac{\varepsilon}{2} \cdot $
Since $ \lim\limits_{x\to +\infty}{\frac{f\left(A\right)-A\ell}{x}}=0 $, there is some $ A_{2}\geq a $ such that $ \left(\forall x\geq A_{2}\right),\ \frac{\left|f\left(A\right)-A\ell\right|}{x}<\frac{\varepsilon}{2} $
For $ x\geq A=\max\left(A_{1},A_{2}\right) $, we have : \begin{aligned} \left|\frac{f\left(x\right)}{x}-\ell\right|&=\frac{1}{x}\left|\int_{A}^{x}{\left(f'\left(t\right)-\ell\right)\mathrm{d}t}+f\left(A\right)-A\ell\right|\\ &\leq\frac{1}{x}\int_{A}^{x}{\left|f'\left(t\right)-\ell\right|\mathrm{d}t}+\frac{\left|f\left(A\right)-A\ell\right|}{x}\\\left|\frac{f\left(x\right)}{x}-\ell\right|&\leq\frac{1}{x}\int_{A}^{x}{\frac{\varepsilon}{2}\,\mathrm{d}t}+ \frac{\varepsilon}{2}= \varepsilon-\frac{A\varepsilon}{2x}\leq \varepsilon\end{aligned}
Meaning, $ \lim\limits_{x\to +\infty}{\frac{f\left(x\right)}{x}}=\ell \cdot $
Now since $ \lim\limits_{x\to +\infty}{f\left(x\right)}=\lim\limits_{x\to +\infty}{\underbrace{x}_{\underset{x\to +\infty}{\longrightarrow}+\infty}\underbrace{\left(\frac{f\left(x\right)}{x}\right)}_{\underset{x\to +\infty}{\longrightarrow}\ell}} $, we should separate the cases to determine, according to the values of the constant $ \ell $, the value of $ \lim\limits_{x\to +\infty}{f\left(x\right)} $ :
If $ \ell >0 $ the limit would be $ +\infty $, if $ \ell <0 $ the limit would be $ -\infty $, and when $ \ell =0 $ it gives an indeterminate form, so we can not conclude anything about its value.
A: Let $\epsilon>0$ be given and we assume that $f'(x) \to l$ as $x\to\infty $. Here $l$ is a real number. The proof below works with minor modifications when $l=\pm\infty $.
By definition of limit there is a positive number $a$ such that $$|f'(x) - l|<\frac {\epsilon} {3}\tag{1}$$ whenever $x>a$. Next consider the algebraic identity $$\frac{f(x)} {x}-l =\left\{\left(1-\frac{a}{x}\right)\cdot\frac{f(x) - f(a)} {x-a}-l\right\} +\frac{f(a)} {x} \tag{2}$$ (the above is just plain algebra with no magic involved). And then our strategy is to ensure that both terms in $(2)$ are small. Clearly the second term is less than $\epsilon /3$ in absolute value if $x>\dfrac{3(|f(a)|+1)}{\epsilon}=b$. Using mean value theorem the first term in $(2)$ can be written as $$\left(1-\frac{a}{x}\right)f'(\xi)-l=f'(\xi) - l-\frac{af'(\xi)} {x} \tag{3}$$ for some $\xi\in(a, x) $. The absolute value of last term in above equation does not exceed $a(|l|+(\epsilon/3))/x$ and hence does not exceed $\epsilon /3$ if $$x>\frac{3a(|l|+(\epsilon/3))}{\epsilon}=c$$ Let $x>N=\max(a, b, c) $ and then we have via triangle inequality applied on $(2)$ and $(3)$ $$\left|\frac{f(x)} {x} - l\right|\leq |f'(\xi) - l|+a\cdot\frac{|f'(\xi) |} {x}+\frac{|f(a) |} {x} \tag{4}$$ where $\xi$ is some number between $a$ and $x$. The way $a,b,c, N$ are defined each term in right side of $(4)$ is less than $\epsilon /3$ when $x>N$ and therefore the left hand side of $(4)$ is less than $\epsilon$ whenever $x>N$. By definition of limit we have $$\lim\limits _{x\to\infty} \dfrac{f(x)} {x} =l$$ as desired.
Note that given $l$ and $\epsilon$ we first determine $a$. Then using $a$ and $\epsilon$ we determine $b$. And $c$ is based on $a,l, \epsilon $. Since $l$ is fixed as a part of question, the numbers $a, b, c, N$ depend only on $\epsilon $. 
