If $\lim\limits_{x\to\infty} xf(x) = L$, then $\lim\limits_{x\to\infty} f(x) =0$ [duplicate]

Show that if $f: (a,\infty) \rightarrow \mathbb R$ such that $$\lim_{x\to \infty} xf(x) = L$$ where $L \in \mathbb R,$ then $$\lim_{x\to \infty} f(x) = 0.$$

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$\displaystyle\lim_{x\to\infty}\dfrac{1}{x}=0$

$\displaystyle\lim_{x\to\infty}xf(x)=L$

$0\times L=\displaystyle\lim_{x\to\infty}\dfrac{1}{x}\times\lim_{x\to\infty}xf(x)=\lim_{x\to\infty}\dfrac{1}{x}\cdot xf(x)=\lim_{x\to\infty}f(x)$

• +1 This answer is the only one so far that is not assuming the existence of $\lim f(x)$ beforehand. – Hagen von Eitzen Oct 23 '13 at 13:07

Hint. If $\lim_{x\to\infty} f(x)\neq 0$ then there is a $c>0$ and increasing sequence $(a_n)$ such that $\lim_{n\to\infty} a_n=+\infty$ and $|f(a_n)|>c$ for all $n$. In that cases, $a_nf(a_n)$ converges?

$$lim_{x \rightarrow \infty} xf(x) = lim_{x \rightarrow \infty} x \cdot lim_{x \rightarrow \infty} f(x)$$

Since $lim_{x\rightarrow \infty} = \infty$, if $$lim_{x \rightarrow \infty} f(x) = k \ne 0$$, where $k$ is a real number, then $$lim_{x \rightarrow \infty} xf(x) = \pm \infty \ne L$$ the $\pm$ sign depends on $k$.