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Which is the solution of the limit,

$$ \lim_{x \rightarrow \infty} x\sin \left(\frac{1}{x}\right) $$

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    $\begingroup$ Use substitution: $u=\frac{1}{x}$ and note that $x\to\infty\Rightarrow u\to0$ $\endgroup$ – cjferes Nov 5 '14 at 20:19
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You can do this substitution: $$x=\frac{1}{u},\ u=\frac{1}{x},\ x\rightarrow\infty\Rightarrow u\rightarrow0$$ Hence, your limit changes to: $$\lim_{u\rightarrow0}\frac{1}{u}\sin(u)=\lim_{u\rightarrow0}\frac{\sin(u)}{u}$$ A very known limit.

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Put $t= \dfrac 1x \iff x = \dfrac 1t$. And note that as $x\to \infty,\;t\to 0^+$.

$$\lim_{x\to \infty} x \sin \left(\frac 1x\right) = \lim_{t\to 0^+} \dfrac{\sin(t)}{t}$$

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Since $\sin \frac1x \sim \frac1x$ as $x\to \infty$, then $\lim_{x\to \infty}x\sin \frac1x = \lim_{x\to \infty}x \frac1x=1 $

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