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.


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 $\lim_{x\to\infty} xf'(x)=0$.

  • $\begingroup$ This is a respectable use of L'Hospital's Rule. +1 $\endgroup$ – Paramanand Singh Feb 4 '18 at 16:47
  • $\begingroup$ This is essentially your argument, but yours is much more streamlined :) $\endgroup$ – pisco Feb 4 '18 at 16:50
  • $\begingroup$ @pisco125 Why does the first step imply that the $limx→∞ f'(x) = 0 $? $\endgroup$ – Jeremy Gavriel Feb 5 '18 at 2:09
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    $\begingroup$ @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. $\endgroup$ – Paramanand Singh Feb 5 '18 at 2:28
  • $\begingroup$ $\sin(x)/x$ tends to 0 as $x$ tends to infinity. However, $\sin(x)$ does not. $\endgroup$ – Robert Wolfe Feb 5 '18 at 3:38

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 $x\to\infty$.


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