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Let $f ∈ C^2$ and $f:(0,∞)→R$ be a function such that $\underset{x \rightarrow \infty}{\lim}xf(x)=0$ and $\underset{x \rightarrow \infty}{\lim}xf''(x)=0$. Prove that $\underset{x \rightarrow \infty}{\lim}xf'(x)=0$.

I somehow feel that this problem needs to be solved using Taylor expansion (taylor series) since it involves derivatives. However, I do not know how to show it.

Jeremy Gavriel
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2 Answers2

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I use the following version of Lopital's rule: if $f(x),g(x)$ are differentiable on $(0,\infty)$, with $|g(x)|\to \infty $ as $x\to \infty$. Then $$\lim_{x\to \infty} \frac{f'(x)}{g'(x)} = L \implies \lim_{x\to \infty} \frac{f(x)}{g(x)} = L$$

Note that no assumption is imposed on $f(x)$. The proof is reminiscent to that of Stolz's theorem.

We use the lopital's rule on the function $$\frac{e^x f'(x)-e^xf(x)}{e^x}$$

To obtain $$\lim_{x\to\infty} \frac{(e^x f'(x)-e^x f(x))'}{(e^x)'} = \lim_{x\to\infty} \frac{e^xf''(x)-e^xf(x)}{e^x} = \lim_{x\to\infty} (f''(x)-f(x))=0$$ Hence $$\lim_{x\to\infty}\frac{e^x f'(x)-e^xf(x)}{e^x} =0=\lim_{x\to\infty}(f'(x)-f(x))$$

This implies $\lim_{x\to\infty} f'(x) = 0$.

Now use lopital's rule on $$\frac{xe^x f'(x)-xe^xf(x)}{e^x}$$ To obtain $$\begin{aligned}\lim_{x\to\infty} \frac{(xe^x f'(x)-xe^x f(x))'}{(e^x)'} &= \lim_{x\to\infty} \frac{e^xf'(x)+xe^xf''(x)-e^x f(x)-xe^x f(x)}{e^x} \\ &= \lim_{x\to\infty} (f'(x)+xf''(x)-f(x)-xf(x))=0 \end{aligned}$$ where we used result obtained above. Hence $$\lim_{x\to\infty}\frac{xe^x f'(x)-xe^xf(x)}{e^x} =0=\lim_{x\to\infty}(xf'(x)-xf(x))$$ This implies $\lim_{x\to\infty} xf'(x)=0$.

pisco
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  • This is a respectable use of L'Hospital's Rule. +1 – Paramanand Singh Feb 04 '18 at 16:47
  • This is essentially your argument, but yours is much more streamlined :) – pisco Feb 04 '18 at 16:50
  • @pisco125 Why does the first step imply that the $limx→∞ f'(x) = 0 $? – Jeremy Gavriel Feb 05 '18 at 02:09
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    @JeremyGavriel: note that $f$ already tends to $0$ (because $xf$ does so) and it is proved that $f'-f$ tends to $0$ and thus their sum $(f'-f) +f=f'$ also does the same. – Paramanand Singh Feb 05 '18 at 02:28
  • $\sin(x)/x$ tends to 0 as $x$ tends to infinity. However, $\sin(x)$ does not. –  Feb 05 '18 at 03:38
  • @Robert Yes, that is true, but in this case it's $x$ multiplied by a function (in your example, maybe $x * sinx$ is the better argument. You raise a good point though... – Jeremy Gavriel Feb 05 '18 at 03:42
  • @pisco125 How can we be sure that the first function where you used L'Hopital is an indeterminate form? Is it enough to just say that the function does not go to zero faster than the speed where $e^x$ goes to infinity? – Jeremy Gavriel Feb 05 '18 at 06:49
  • @Robert I used the following version of Lopital's rule: if $f(x),g(x)$ are differentiable on $(0,\infty)$, with $|g(x)|\to \infty $ as $x\to \infty$. Then $$\lim_{x\to \infty} \frac{f'(x)}{g'(x)} = L \implies \lim_{x\to \infty} \frac{f(x)}{g(x)} = L$$ Note that no assumption is imposed on $f(x)$. – pisco Feb 05 '18 at 10:27
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The theorem mentioned in this answer is the key here. Since $f, f''$ both tend to $0$ the derivative $f' $ also tends to $0$. Again note that if $g(x) =xf(x) $ then $g''(x) =xf''(x) +2f'(x) $ and thus $g, g''$ both tend to $0$. By the theorem in linked answer $g' =xf'+f$ tends to $0$ and we are done.

It is also clear from the above that the result holds under the weaker assumption that $xf''(x)$ is bounded as $x\to\infty$.

Paramanand Singh
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