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 $0$.
Answer
Note that
$$-\frac{\sqrt x}{x+1}\le \frac{\sqrt x\cos(x-x^2)}{x+1}\le\frac{\sqrt x}{x+1}$$
and
$$\frac{\sqrt x}{x+1}\to 0$$
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