Proving $ \lim_{x\to\infty}\frac{\sqrt x\cos(x-x^2)}{x+1} = 0. $ I have to prove that 
$$ \lim_{x\to\infty}\frac{\sqrt x\cos(x-x^2)}{x+1} = 0. $$
I tried squaring both the denominator and numerator to get rid of $\sqrt{x}$ but then $\cos$ becomes $\cos^2$ and I do not know how to solve it/simplify it further. 
I have also tried to use the squeeze theorem with $a_n = 0$, and $$ b_n =\frac{\sqrt x\cos(x-x^2)}{x+1}, $$ but to no avail. I could not find another function that is greater than $b_n$ that has a limit of $0$.
 A: @gimusi shows 'why'.
And this is 'how':
The first thing to note is that cosine remains between $-1$ and $+1$, no matter what its argument is.
The second one is that a linear expression in the denominator grows faster than a square root in the numerator.
Then you can squeeze the cosine with $$\pm\frac{\sqrt x}{x+1} $$ whose absolute value is in turn less than $1/{\sqrt x}$.
A: Note that
$$-\frac{\sqrt x}{x+1}\le \frac{\sqrt x\cos(x-x^2)}{x+1}\le\frac{\sqrt x}{x+1}$$
and
$$\frac{\sqrt x}{x+1}\to 0$$
A: Let $-1\le \cos \left(x-x^2\right)\le 1$, then 
$$\lim _{x\to \infty }\left(\frac{\sqrt{x}\left(-1\right)}{x+1}\right)\le \lim _{x\to \infty }\left(\frac{\sqrt{x}\cos \left(x-x^2\right)}{x+1}\right)\le \lim _{x\to \infty }\left(\frac{\sqrt{x}\cdot1}{x+1}\right)$$
It follows
$$\lim _{x\to \infty }\left(\frac{\sqrt{x}\left(\pm1\right)}{x+1}\right) = \pm\lim _{x\to \infty }\left(\frac{\sqrt{x}}{x+1}\right) = \pm\lim _{x\to \infty }\left(\frac{\frac{1}{\sqrt{x}}}{1+\frac{1}{x}}\right)=\pm\frac{0}{1} = 0$$
