Proving $ \lim_{x\to\infty}\frac{\sqrt x\cos(x-x^2)}{x+1} = 0. $
Note that
$$-\frac{\sqrt x}{x+1}\le \frac{\sqrt x\cos(x-x^2)}{x+1}\le\frac{\sqrt x}{x+1}$$
and
$$\frac{\sqrt x}{x+1}\to 0$$
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 $$\lim _{x\to \infty }\left(\frac{\sqrt{x}\left(\pm1\right)}{x+1}\right) = \pm\lim _{x\to \infty }\left(\frac{\sqrt{x}}{x+1}\right) = \pm\lim _{x\to \infty }\left(\frac{\frac{1}{\sqrt{x}}}{1+\frac{1}{x}}\right)=\pm\frac{0}{1} = 0$$
@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 $1/{\sqrt x}$.