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Let $f$ be a real function continuously differentiable at $\Bbb R$ such that $$\lim_{x\to +\infty}(f(x)+f'(x))=0$$ prove that

$$\lim_{x\to +\infty} f(x)=0$$ I tried tu use exponential function knowing that

$$\frac{d}{dx}f(x)e^x=(f(x)+f'(x))e^x$$ but I got nothing. thanks in advance for an answer or un idea

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  • $\begingroup$ Have you tried to solve the differential equation $f'(x) = -f(x)+ g(x)$, where $g(x)$ is a function, that goes towards $0$ for large $x$? $\endgroup$ Apr 19, 2020 at 20:19
  • $\begingroup$ @HamidMohammad No i did not {}{}{}{} $\endgroup$ Apr 19, 2020 at 20:20
  • $\begingroup$ Thanks a lot a lot a lot. $\endgroup$ Apr 19, 2020 at 20:57

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Could you prove it by contradiction? If $\lim_{x\rightarrow \infty} f'(x) \neq 0$ then $\lim_{x\rightarrow \infty} f(x) = -\lim_{x\rightarrow \infty} f'(x)$ but if $f'>0$ then $f$ is increasing, and if $f'<0$ then $f$ is decreasing - that seems like a contradiction.

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    $\begingroup$ What if $f’$ has no limit? $\endgroup$ Apr 19, 2020 at 21:05

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