Let $l \in\mathbb{R} $ and $ f:(0,\infty) \to \Bbb R $ be a function such that $ \lim_{x\to \infty} xf(x)=l $ .

Prove that $ \lim_{x\to \infty} f(x)=0 $.

Any help would be appreciated - I found this difficult to prove


$\lim\limits_{x\to \infty} xf(x)=l$

$\lim\limits_{x\to \infty} \frac{1}{x}=0$

$\lim\limits_{x\to \infty} \frac{1}{x}xf(x) =\lim\limits_{x\to \infty}f(x)$

$\lim\limits_{x\to \infty} \frac{1}{x}xf(x)=\left(\lim\limits_{x\to \infty} \frac{1}{x}\right)\left(\lim\limits_{x\to \infty} xf(x)\right)=0l=0$

$\boxed{\lim\limits_{x\to \infty}f(x)=0}$



$$\lim_{x\to\infty}f(x)=\frac l{\lim_{x\to\infty} x}$$

  • 1
    $\begingroup$ Who tells you that such a limit exists? $\endgroup$ – Edoardo Lanari Jun 23 '13 at 14:40
  • $\begingroup$ @Lano, isn't it legal to write $$\lim_{x\to\infty}f(x)=\frac l{\lim_{x\to\infty} x}?$$ $\endgroup$ – lab bhattacharjee Jun 23 '13 at 14:43
  • $\begingroup$ Only if $\lim_{x\to\infty}f(x)$ exists, and you have to prove it. $\endgroup$ – Edoardo Lanari Jun 23 '13 at 14:44
  • $\begingroup$ @Lano, I think the question assumes the existence of the limit, for example $$\lim_{x\to0}x\cdot \sin\frac1x=0(l \text{ here} ),$$ but $$\lim_{x\to0} \sin\frac1x$$ does not exist, right? $\endgroup$ – lab bhattacharjee Jun 23 '13 at 15:00
  • 1
    $\begingroup$ There's no need for such an assumption:$\left(\lim\limits_{x\to \infty} \frac{1}{x}\right)\left(\lim\limits_{x\to \infty} xf(x)\right)=0l=0$ $\endgroup$ – Edoardo Lanari Jun 23 '13 at 15:27

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