Can you find $\lim_{x\to0}x^2\sin\left(\frac{1}{x}\right)$ without sandwich theorem? The lesson I'm doing is solving 
$\lim_{x\to0}x^2\sin\left(\dfrac{1}{x}\right)$
with sandwich theorem like so :
$$\displaystyle\lim_{x\to0}-x^2\leq\displaystyle\lim_{x\to0}x^2\sin\left(\dfrac{1}{x}\right)\leq\displaystyle\lim_{x\to 0}x^2$$
When I first look at this though my thought was to use the properties of lim and do this :
$\lim_{x\to0}x^2 = 0$ and times that by $\lim_{x\to0}\sin\left(\dfrac{1}{x}\right)$
And if I'm not an idiot then this says that $0\sin(1/0)$ is going to equal $0$ which is the right answer.
Does that make sense? Instead of bothering with sandwich theorem you could just pull apart the $x^2$ and $\sin(1/x)$, input the $0$ of the limit and not even have to bother with the sine.

I think It make sense. see whatever is in sine function it will always give a value from [-1,1] and if we multiply it by a number which is almost zero we will get zero.
