How to prove that both f(x) and its derivative decay to zero, as x grows to infinity, The problem statement is:
Prove: If $f(x)$ is differentiable for x>0 and 
$$\lim_{x \to \infty}  (f(x)+\frac {df}{dx}(x)) =0$$
and if 
$$\lim_{x \to \infty} f(x)$$ exists, then  
$$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{df}{dx}(x)=0$$
My work:
I don't see how the differentiability assumption on f can be used (or is even needed?), but I have that, for every $\epsilon$>0, there exists M = $max(M_1,M_2)$, such that x>M implies that
$$|f(x)+ \frac{df}{dx}(x)| < \epsilon$$
$$\implies-\epsilon < f(x) +  \frac{df}{dx}(x)  < \epsilon$$
$$\implies -\frac{df}{dx}(x) -\epsilon < f(x)   < -\frac{df}{dx}(x) +\epsilon$$
and since by assumption we know that the limit at infinity for f(x) exists, then the limit must be $-\frac{df}{dx}$, by the last inequality above.  
Now I think the goal is to just show that the limit at infinity of $|\frac{df}{dx}|=0.$  Then we would have the desired equality, by the Squeeze Theorem.
But what can I say about the limit of $|\frac{df}{dx}|$?
Thanks,
 A: With no loss of generality, suppose $\lim_{x \to \infty}f(x) > 0$; then by assumption we have $\lim_{x \to \infty}f'(x) < 0$; this implies that there are some $X \in \mathbb{R}$ and some $l < 0$ such that
$f'(x) < l$ for all $x \geq X$, which in turn implies that for every $x > X$ there is some $X < c < x$ such that
$$
f(x) - f(X) = (x-X)f'(c) < (x-X)l
$$
by the mean-value theorem;
this says that $\lim_{x \to \infty}f(x)$ does not exist, a contradiction.
A: You can force the limits into L'hopital's rule form:
$$\lim_{x\rightarrow\infty}f(x) =\lim_{x\rightarrow\infty} \dfrac{e^x f(x)}{e^x}\underset{(L)}{=}\lim_{x\rightarrow\infty}\dfrac{e^x f(x) + e^x f'(x)}{e^x} = \lim_{x\rightarrow\infty}f(x)+f'(x) = 0 $$
The use of L'hopital's rule is due to the fact that all the limits exist and the 2nd from left limit is of the form of $\dfrac{*}{\infty}$ (L'hopital's rule holds for this cases aswell)

Now we can calculate the limit of the derivative: 
$$\lim_{x\rightarrow\infty}f'(x) = \lim_{x\rightarrow\infty} f'(x) + f(x) - f(x) = \lim_{x\rightarrow\infty}f'(x) + f(x) - \lim_{x\rightarrow\infty} f(x) = 0-0 = 0$$
A: If $f'(x) > M > 0$ for $x > N$, then what can you say about $f(x) = C + \int_N^x f'(x)dx$ as $x \to \infty$ ?
