# I have to evaluate the following without L'Hopital's rule <munder> <mo movablelimits="tru

I have to evaluate the following without L'Hopital's rule
$\underset{x\to \mathrm{\infty }}{lim}x\mathrm{tan}\left(1/x\right)$
I can simplify this to be
$\underset{x\to \mathrm{\infty }}{lim}x\mathrm{sin}\left(1/x\right)$
because
$\underset{x\to \mathrm{\infty }}{lim}\mathrm{cos}\left(1/x\right)=1$
However, after that, I'm totally lost. L'Hopital's rule seems like my only option. Can someone help me out?
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Elias Flores
If you are allowed to use the fact that $\underset{t\to 0}{lim}\frac{\mathrm{sin}\left(t\right)}{t}=1$, then setting $t=\frac{1}{x}$ yields
$\underset{x\to \mathrm{\infty }}{lim}x\mathrm{sin}\left(\frac{1}{x}\right)=\underset{x\to \mathrm{\infty }}{lim}\frac{\mathrm{sin}\left(\frac{1}{x}\right)}{\frac{1}{x}}=\underset{t\to 0}{lim}\frac{\mathrm{sin}\left(t\right)}{t}=1$