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

$$|f(x)+ \frac{df}{dx}(x)| &lt; \epsilon$$
$$\implies-\epsilon &lt; f(x) +  \frac{df}{dx}(x)  &lt; \epsilon$$
$$\implies -\frac{df}{dx}(x) -\epsilon &lt; f(x)   &lt; -\frac{df}{dx}(x) +\epsilon$$

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,
With no loss of generality, suppose $\lim_{x \to \infty}f(x) &gt; 0$; then by assumption we have $\lim_{x \to \infty}f'(x) &lt; 0$; this implies that there are some $X \in \mathbb{R}$ and some $l &lt; 0$ such that
$f'(x) &lt; l$ for all $x \geq X$, which in turn implies that for every $x &gt; X$ there is some $X &lt; c &lt; x$ such that
$$
f(x) - f(X) = (x-X)f'(c) &lt; (x-X)l
$$
by the mean-value theorem;
this says that $\lim_{x \to \infty}f(x)$ does not exist, a contradiction.
If $f'(x) &gt; M &gt; 0$ for $x &gt; N$, then what can you say about $f(x) = C + \int_N^x f'(x)dx$ as $x \to \infty$ ?