If $f:\mathbb R \to \mathbb R^+$ is a differentiable function and $g(x)=e^x \cdot f(x)$ such that $\displaystyle\lim_{x \to \infty} (f(x)+f'(x)) = 0$, find the limit: $$\lim\limits_{x, y \to \infty} \frac{g(x)-g(y)}{e^x-e^y}$$
I know from this that $f(x)$ goes to zero as $x$ goes to infinity. Can I get hints on how to find the desired limit?
As @PhoemueX has pointed out, we can use CMVT to solve this. Verify that the assumptions to apply CMVT hold. Now see that $\exists \ z \in (x,y)$ (assuming WLOG $x < y$) such that
$$ \frac{g(x) - g(y)}{e^x - e^y} = \frac{g'(z)}{e^z} = \frac{e^z(f(z) + f'(z))}{e^z} = f(z) + f'(z).$$
Apply limits on both sides to get $$\lim_{x,y \to \infty} \frac{g(x) - g(y)}{e^x - e^y} = \lim_{\substack{z \ \in \ (\min(x,y),\max(x,y)) \\ x,y \ \to \ \infty}} (f(z) + f'(z)) = 0.$$