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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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    $\begingroup$ Hello, welcome to Math.SE. Thank you for your question! It is good practice on this site to add a bit of information on the context your question came up in, and to share your own work on it. It's also fine if you state that you're completely lost -- the information is helpful for answerers to gauge their answer on. For more information on asking a good question on this site, see here. $\endgroup$ – Lord_Farin Oct 23 '13 at 12:58
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$\displaystyle\lim_{x\to\infty}\dfrac{1}{x}=0$

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

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

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

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

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 $k$.

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