# Show that $\displaystyle \lim_{x\to \infty} f(x) =0$

Let $$f:\mathbb{R}^+ \to \mathbb{R}$$ be a differentiable function. Consider the function $$g(x) = e^x f(x)$$. If $$\lim_{x\to \infty} \left(f(x) + f'(x) \right) = 0,$$ then show that $$\displaystyle \lim_{x\to \infty} f(x) = 0$$.

Note that $$g'(x) = e^x \left(f(x) + f'(x)\right) \implies \lim_{x\to \infty} \dfrac{g'(x)}{e^x} = 0.$$ We need to evaluate the following limit: $$\lim_{x\to \infty} f(x) = \lim_{x\to \infty} \dfrac{g(x)}{e^x}.$$ If we know that the $$\lim_{x\to \infty} g(x) = \infty$$, then we can use the L'Hosptial rule to conclude that $$\lim_{x\to \infty}f(x) = 0$$. But we do not know if $$\lim_{x\to \infty}g(x) = \infty$$.

Any hint(s) will be appreciated.

Thanks in advance!

## 1 Answer

If there exists a constant $$C> 0$$ such that $$|g(x)| \le C$$ for every $$x$$ sufficiently large, then $$|f(x)| \le \frac{C}{e^x} \to 0 \quad \text{as } x \to \infty.$$ This gives you the result for this $$g(x)$$. For the case where $$g$$ is not bounded at infinity, as you said, you can apply de L'Hospital rule.