Proving a limit equals zero I have to prove, without L'hopital rule, the following limit:
$$\lim_{x \to \infty}\sqrt{x} \sin \frac{1}{x} =0$$ 
I tried doing a variable change, setting $t=\frac{1}{x}$ and reaching the following: $$\lim_{t\to 0} \sqrt{\frac{1}{t}} \sin t $$
But I can't prove neither. Tried the second version with the squeeze theorem, but I can't prove the limit.
Thanks!

$$\lim_{t \to 0} \frac{1}{\sqrt{t}} \sin(t) = \lim_{t \to 0} \frac{1}{t} \sin(t^2) = \lim_{t \to 0} \frac{\sin t^2 - \sin 0^2}{t} = f'(0)$$
where $f(x) = \sin x^2$.
By the chain rule, $f'(x) = 2x \cos x^2$, which is $0$ at $x=0$.
