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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!

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1 Answer 1

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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.

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