# Showing that if $\lim_{x\to\infty}xf(x)=l,$ then $\lim_{x\to\infty}f(x)=0.$

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

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

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

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

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

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

HINT:

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

• Who tells you that such a limit exists? Commented Jun 23, 2013 at 14:40
• @Lano, isn't it legal to write $$\lim_{x\to\infty}f(x)=\frac l{\lim_{x\to\infty} x}?$$ Commented Jun 23, 2013 at 14:43
• Only if $\lim_{x\to\infty}f(x)$ exists, and you have to prove it. Commented Jun 23, 2013 at 14:44
• @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? Commented Jun 23, 2013 at 15:00
• There's no need for such an assumption:$\left(\lim\limits_{x\to \infty} \frac{1}{x}\right)\left(\lim\limits_{x\to \infty} xf(x)\right)=0l=0$ Commented Jun 23, 2013 at 15:27