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!
 A: 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
$$\lim_{x\to\infty}\sqrt{x}\sin\dfrac1x=0$$
A: $$\lim_{t \to 0} \frac{1}{\sqrt{t}} \sin(t) = \lim_{t \to 0} \frac{1}{t} \sin(t^2) = \lim_{t \to 0} \frac{\sin t^2 - \sin 0^2}{t} = f'(0)$$
where $f(x) = \sin x^2$.
By the chain rule, $f'(x) = 2x \cos x^2$, which is $0$ at $x=0$.
A: 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
$$
\lim_{x\to\infty} \frac{1}{\sqrt{x}}=0\implies \lim_{x\to\infty} \sqrt{x}\sin\frac{1}{x}=0
$$
A: $$\lim_{t\to0}\frac{\sin t}{\sqrt t}=\lim_{t\to0}\frac{\sin t}t\sqrt t=\lim_{t\to0}\frac{\sin t}t\cdot\lim_{t\to0}\sqrt t=1\cdot0.$$
A: $$\sin {t}=\sum_{n=0}^{+\infty}\frac{(-1)^nt^{2n+1}}{(2n+1)!}=t-\frac16t^3+\frac1{120}t^5-\dots=\sqrt{t}\left(\sqrt{t}-\frac16t^{3-\frac12}+\frac1{120}t^{5-\frac12}-\dots\right)$$
