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!

first we need a bound on $\sin$:
$$
\left|\sin(x)\right|=\left|\sin(x)-\sin(0)\right|=\left|\int_0^x \cos(t)\,\textrm{dt}\right|\le\left|\int_0^x |\cos(t)|\,\textrm{dt}\right|\le\left|\int_0^x 1\,\textrm{dt}\right|=|x|
$$
so now we have a useful inequality:
$$
|\sin(x)|\le|x|\,\,\,\,\forall x\in\mathscr{R}
$$
Bearing in mind that $x>0$ we can apply it to your limit this way:
$$
0\le\left|\sqrt{x}\sin\frac{1}{x}\right|=\sqrt{x}\left|\sin\frac{1}{x}\right|\le\sqrt{x}\frac{1}{x}=\frac{1}{\sqrt{x}}
$$
now by the squeeze theorem
$$
\lim_{x\to\infty} \frac{1}{\sqrt{x}}=0\implies \lim_{x\to\infty} \sqrt{x}\sin\frac{1}{x}=0
$$
