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This question already has an answer here:

Suppose that $f:[0,\infty)\to\Bbb R$ is differentiable, and $\lim_{x\to\infty}(f(x)+f'(x))=0$. Prove that $\lim_{x\to\infty}f(x)=0$.

I tried to show that $\lim_{x\to\infty}f(x)\neq0\rightarrow \lim_{x\to\infty}(f(x)+f'(x))\neq0$

If $\lim_{x\to\infty}f(x)\neq0, \exists e>0, \forall N\in\Bbb N, \exists x_1, x_2, ...>N,|f(x_i)|\ge e$.

So that the possibility for $\lim_{x\to\infty}(f(x)+f'(x))=0$ is only when $f(x_i)+f'(x_i)=0$ is true for any $i$.

So if $f(x_1)\gt0,$ it must be decreasing, to less than $e$

But there exists $x_2\gt x_1 s.t. |f(x_2)|\ge e$

In conclusion, $f(x)$ has to decrease when $x$ is large enough and $f(x)\ge e$ but there must exists infinitely many points whose function value is no less than $e$. But it cannot happen.

Is my idea of proof valid?

I don't know how to formally write my idea...please teach me..

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marked as duplicate by Hans Lundmark, user296602, Henrik, user99914, Namaste Jul 16 '16 at 21:21

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By L'H we have

$$\lim_{x \to \infty} \frac{f(x) e^x}{e^x}=\lim_{x \to \infty} \frac{(f(x)+f'(x)) e^x}{e^x}=0$$

Note that L'H can be applied as we are in the case $\frac{\mbox{something}}{\infty}$.

Alternately Apply Cauchy's Mean Value Theorem to $g(x)=e^xf(x)$ and $h(x)=e^x$.

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    $\begingroup$ Sorry, but L'Hopital can only be applied if $\frac{\infty}{\infty}$ or $\frac{0}{0}$. See en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule $\endgroup$ – user3154270 Jul 16 '16 at 12:30
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    $\begingroup$ @user3154270 That is a common misconception about LH, in the so called $\frac{\infty}{\infty}$ case you only need the denominator to go to $\infty$. $\endgroup$ – N. S. Jul 16 '16 at 12:34
  • $\begingroup$ See here: math.stackexchange.com/questions/891361/… $\endgroup$ – smcc Jul 16 '16 at 12:35
  • $\begingroup$ @user3154270 This is quoted from the link you provided: " In the second case, the hypothesis that f diverges to infinity is not used in the proof (see note at the end of the proof section); thus, while the conditions of the rule are normally stated as above, the second sufficient condition for the rule's procedure to be valid can be more briefly stated as $\lim_{x \to c} |g(x)|=\infty$." $\endgroup$ – N. S. Jul 16 '16 at 12:35
  • $\begingroup$ Anyway, if $\lim_{x\to\infty} f(x) = c < \infty$, then $\lim_{x\to\infty} f(x)e^x = \infty$, so we are actually in the case $\frac{\infty}{\infty}$. So I think this proof seems to be correct $\endgroup$ – user3154270 Jul 16 '16 at 12:57

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