squeezee theorem

A small question to start with
Q) Find the limit as x → 0 of the following function on R

\phi (x) = \left\{\begin{matrix} xsin\frac{1}{x} ( x\neq 0\\ 0(x =0)\end{matrix}\right.}

solution :
since \left| sin\frac{1}{x}\right| \leq 1
it follows that
0\leq \left| \phi (x)\right| = \left|xsin\frac{1}{x} \right|\leq \left| x\right|
with f(x) = 0, g(x) = |x|
we have \lim_{x\rightarrow 0}f(x) =0,\lim_{x\rightarrow 0}g(x) =0
and so by sandwich theorem \lim_{x\rightarrow 0}\phi (x) =0
hence \lim_{x\rightarrow 0}\left|xsin\frac{1}{x} \right| =0

For xsin(1x), I thought we usually prove it like the sine function provides values between -1 and 1 and x -> 0 implies
0 * something finite = 0.
But that's a nice way to prove it too!