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!

$$\lim_{t\to0}\frac{\sin t}{\sqrt t}=\lim_{t\to0}\frac{\sin t}t\sqrt t=\lim_{t\to0}\frac{\sin t}t\cdot\lim_{t\to0}\sqrt t=1\cdot0.$$
