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Let $f$ be differentiable on $(0,\infty)$. Show that $\lim\limits_{x\to \infty}(f(x)+f'(x))=0$, then $\lim\limits_{x\to \infty}f(x)=0$

My attempt:

When both $f(x)$ and $f'(x)$ $\to 0$ when $x\to \infty$ then the problem is trivial. But, the problem is I cannot do anything more than the trivial case. I definitely realise that if $\lim\limits_{x\to \infty}f(x)=0$ then $\lim\limits_{x\to \infty}(f(x)+f'(x))=0$ is true. But, I cannot really prove the converse. Please help. Thank you.

Swadhin
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2 Answers2

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Hint: Apply L'Hospital's rule to $$\frac{e^xf(x)}{e^x}.$$

Eclipse Sun
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  • The exponent of $e$ is too small for me to see. – Swadhin Mar 09 '15 at 05:00
  • Right. I realise you are very correct! Thank you. – Swadhin Mar 09 '15 at 05:05
  • There is subtlety to the logic here though; notice that in using L'Hospital's rule, we must assume that $\lim_\limits{x\to\infty} f(x)\neq 0$. – rnrstopstraffic Mar 09 '15 at 05:29
  • L'Hospital's Rule requires that both the numerator and denominator terms approach, in this case, infinity. You're trying to prove that $f$ goes to zero. If it does, then how does one know that $e^xf(x)$ still goes to infinity? Or perhaps $\lim_{x\to \infty} e^xf(x)$ does not exist (e.g., $f(x)=\sin(x)$). – Mark Viola Mar 09 '15 at 05:49
  • @Dr.MV L'Hospital's Rule does NOT require both terms approach to infinity, though its usual form is what you mentioned. The same is true for a related theorem of Stolz. By the way, as for $\frac{0}{0}$, BOTH terms should approach to zero! – Eclipse Sun Mar 09 '15 at 06:51
  • @rnrstopstraffic See the comment above. We only need the denominator approaches to infinity. – Eclipse Sun Mar 09 '15 at 06:54
  • @EclipseSun That's simply not true. Consider $\lim\limits_{x\to\infty}\frac{\sin (x)}{x}=0$. The denominator approaches infinity, but this clearly doesn't work if you differentiate a la L'Hospital. – rnrstopstraffic Mar 10 '15 at 05:14
  • @rnrstopstraffic Consider $\lim\limits_{x\to\infty}\frac{x-\sin (x)}{x}=0$. Be careful when using L'Hospital. – Eclipse Sun Mar 10 '15 at 06:53
  • @EclipseSun That's exactly the point. We need to be careful when using L'Hospital. I gave an example where the denominator approaches $\infty$ but L'Hospital's rule does not apply. Your example is a perfectly good example of the indeterminate form $\frac{\infty}{\infty}$ to which nobody will object. – rnrstopstraffic Mar 10 '15 at 14:20
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$$\lim_{x\to \infty}f(x)=\lim_{x\to \infty}\frac{e^xf(x)}{e^x}=\lim_{x\to \infty}(f(x)+f'(x))=0$$ I used L'hosptal rule.

Faye Tao
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  • Lorence. L'Hospital's Rule requires that both the numerator and denominator terms approach, in this case, infinity. You're trying to prove that $f$ goes to zero. If it does, then how does one know that $e^xf(x)$ still goes to infinity? Or perhaps $\lim_{x\to \infty} e^xf(x)$ does not exist (e.g., $f(x)=\sin(x)$). – Mark Viola Mar 09 '15 at 05:50
  • @Dr.MV Sorry to say that L'Hospital's Rule doesn't require both the numerator and denominator terms approach infinity,we only need denominator terms approaches infinity and can't be zero. – Faye Tao Mar 09 '15 at 08:55