Originally Posted by

**Chris L T521** This is what I would do:

$\displaystyle \lim_{x\to 0}\cos\left(\frac{1}{x}\right)\sin x=\lim_{x\to 0}{\color{red}\frac{x}{x}}\cdot\cos\left(\frac{1}{ x}\right)\sin x=\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)\frac{\sin x}{x}$

Now, recall that $\displaystyle \lim_{x\to c}f(x)g(x)=\lim_{x\to c}f(x)\cdot\lim_{x\to c}g(x)$

So we have $\displaystyle \lim_{x\to 0}x\cos\left(\frac{1}{x}\right)\frac{\sin x}{x}=\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)\cdot\lim_{x\to 0}\frac{\sin x}{x}$

We should familiarize ourselves with two special limits:

$\displaystyle \lim_{x\to 0}\frac{\sin x}{x}=1$ and $\displaystyle \lim_{x\to \infty}\frac{\cos x}{x}=0$

We need to incorporate these somehow...

We can first start off by simplifying this:

$\displaystyle \lim_{x\to 0}x\cos\left(\frac{1}{x}\right)\cdot\underbrace{\l im_{x\to 0}\frac{\sin x}{x}}_1=\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)\cdot1=\lim_{x\to 0}x\cos\left(\frac{1}{x}\right)$

Now if you make a substitution, say $\displaystyle z=\frac{1}{x}\implies x=\frac{1}{z}$, we see that as $\displaystyle x\to 0$, $\displaystyle z\to\infty$

Thus, $\displaystyle \lim_{x\to 0}x\cos\left(\frac{1}{x}\right)=\lim_{z\to\infty}\ frac{\cos z}{z}=0$

Thus, $\displaystyle \lim_{x\to 0}\cos\left(\frac{1}{x}\right)\sin x=\color{red}\boxed{0}$

Does this make sense?

--Chris