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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..

Arbitrary
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1 Answers1

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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 $h(x)=e^x$.

N. S.
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    Sorry, but L'Hopital can only be applied if $\frac{\infty}{\infty}$ or $\frac{0}{0}$. See https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule – user3154270 Jul 16 '16 at 12:30
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    @user3154270 That is a common misconception about LH, in the so called $\frac{\infty}{\infty}$ case you only need the denominator to go to $\infty$. – N. S. Jul 16 '16 at 12:34
  • See here: http://math.stackexchange.com/questions/891361/how-does-the-l-hopital-rule-work-when-numerator-neq-infty – smcc Jul 16 '16 at 12:35
  • @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 $\lim_{x \to c} |g(x)|=\infty$." – N. S. Jul 16 '16 at 12:35
  • 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 – user3154270 Jul 16 '16 at 12:57
  • @user3154270 Again, read that link. And no, your observation does not solve the problem as you could have $c=0$ or the limit could not exist. – N. S. Jul 16 '16 at 13:00
  • @user3154270 Anyhow if you don't want to use the more general version of LH, I suggested a second proof based on CMVT. All you need to do is apply CMVT and get the claim for free. – N. S. Jul 16 '16 at 13:02
  • Well, if you have $c=0$, then you are done. Because that is what you wanted to show. – user3154270 Jul 16 '16 at 13:05
  • @user3154270 And if the limit doesn't exist? – N. S. Jul 16 '16 at 13:06
  • Assume the limit does not exist, then then you get $c=\infty$ or $c=-\infty$. So you could apply L'Hopital and get a contradiction. – user3154270 Jul 16 '16 at 13:12
  • @user3154270 The limit doesn't exist is not the same thing as being infinite. Just look at $\lim_{x \to \infty} \sin(x)$. And again, you are going to a great deal to try to eliminate an issue which does not exist. Why? – N. S. Jul 16 '16 at 13:14
  • Sorry, my intention is not to make you solution bad. I think it is a very nice one. There are just some details which are interesting for me. – user3154270 Jul 16 '16 at 13:22