# What is the correct solution for the limit <munder> <mo movablelimits="true" form="prefix">

What is the correct solution for the limit $\underset{x\to \mathrm{\infty }}{lim}x\mathrm{sin}\frac{1}{x}$?
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Turynka2f
$\underset{x\to \mathrm{\infty }}{lim}x\mathrm{sin}\frac{1}{x}=\underset{x\to \mathrm{\infty }}{lim}\frac{\mathrm{sin}\left(\frac{1}{x}\right)}{\frac{1}{x}}=\underset{u\to 0}{lim}\frac{\mathrm{sin}\left(u\right)}{u}=1$
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Garrett Black
The first line of reasoning is clearly incorrect: let f(x)=x, $g\left(x\right)=\left({x}^{2}+1{\right)}^{-1}$. Then
$\underset{x\to \mathrm{\infty }}{lim}f\left(x\right)=\mathrm{\infty }$
and $g\left(x\right)\in \left(0,1\right]$ is clearly bounded on $\mathbb{R}$. But
$\underset{x\to \mathrm{\infty }}{lim}f\left(x\right)g\left(x\right)=\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x+1/x}=0.$