# How to evaluate $\lim_{x\to -\infty} x \sin \left(\frac{1}{x}\right)\quad?$

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

My attempt

As x goes to $-\infty$.

-$1/\infty$ is 0, so

$\sin (0)=0$

$-\infty \sin (0)= \infty (0)$

Well I don't know how to evaluate infinity multiplied by zero?

• $\infty\cdot0$ is an "indeterminate form," which means that the answer depends on the limit. – Akiva Weinberger Sep 19 '14 at 16:31

Substitute $u = \frac{1}{x}$. As $x \to -\infty, u \to 0$.

Now, the limit becomes $\lim\limits_{u\to 0}\frac{\sin u}{u}$. Does this seem familiar?

• You're welcome! – taninamdar Sep 19 '14 at 16:07

Knowing that $$\lim_{x \rightarrow 0} \frac{\sin{x}}{x}=1$$ we have the following:

$\displaystyle{\lim_{x \rightarrow - \infty} x \sin{( \frac{1}{x})} =\lim_{x \rightarrow - \infty} \frac{\sin{( \frac{1}{x})}}{\frac{1}{x}}=(*) }$

We set $u=\frac{1}{x}$, so when $x \rightarrow -\infty$ then $u \rightarrow 0$.

Therefore, $$(*)=\lim_{u \rightarrow u} \frac{\sin{u}}{u}=1$$

Hint : $-1 \le sin{\frac 1x} \le 1$ so for any value of x it lies in that range according to the domain of sine function.