# Let $f$ be differentiable on $(0,\infty)$.Show that $\lim\limits_{x\to \infty}(f(x)+f'(x))=0$,then $\lim\limits_{x\to \infty}f(x)=0$

Let $f$ be differentiable on $(0,\infty)$. Show that $\lim\limits_{x\to \infty}(f(x)+f'(x))=0$, then $\lim\limits_{x\to \infty}f(x)=0$

My attempt:

When both $f(x)$ and $f'(x)$ $\to 0$ when $x\to \infty$ then the problem is trivial. But, the problem is I cannot do anything more than the trivial case. I definitely realise that if $\lim\limits_{x\to \infty}f(x)=0$ then $\lim\limits_{x\to \infty}(f(x)+f'(x))=0$ is true. But, I cannot really prove the converse. Please help. Thank you.

#### Solutions Collecting From Web of "Let $f$ be differentiable on $(0,\infty)$.Show that $\lim\limits_{x\to \infty}(f(x)+f'(x))=0$,then $\lim\limits_{x\to \infty}f(x)=0$"

Hint: Apply L’Hospital’s rule to $$\frac{e^xf(x)}{e^x}.$$

$$\lim_{x\to \infty}f(x)=\lim_{x\to \infty}\frac{e^xf(x)}{e^x}=\lim_{x\to \infty}(f(x)+f'(x))=0$$
I used L’hosptal rule.