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!
See that
$|\sin x|\leq x$ while $x\to0$ and here $\dfrac1x\to0$ so 
$$-\dfrac1x\leq\sin\dfrac1x\leq\dfrac1x$$
multiply RH and LH by $\sqrt{x}$ gives us
$$-\sqrt{x}\dfrac1x\leq\sqrt{x}\sin\dfrac1x\leq\sqrt{x}\dfrac1x$$
or
$$-\dfrac{1}{\sqrt{x}}\leq\sqrt{x}\sin\dfrac1x\leq\dfrac{1}{\sqrt{x}}$$
finally the sandwich theorem concludes that
$$\lim_{x\to\infty}\sqrt{x}\sin\dfrac1x=0$$
$$\sin {t}=\sum_{n=0}^{+\infty}\frac{(-1)^nt^{2n+1}}{(2n+1)!}=t-\frac16t^3+\frac1{120}t^5-\dots=\sqrt{t}\left(\sqrt{t}-\frac16t^{3-\frac12}+\frac1{120}t^{5-\frac12}-\dots\right)$$