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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 $0$.

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  • $\begingroup$ Based on intuition, the $\cos$ part doesn't play any role in how the magnitude changes, so it is basically just between $\sqrt{x}$ and $x$, which one of them grows more. $\endgroup$ – IllidanS4 Mar 27 '18 at 20:46
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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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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 $$\lim _{x\to \infty }\left(\frac{\sqrt{x}\left(\pm1\right)}{x+1}\right) = \pm\lim _{x\to \infty }\left(\frac{\sqrt{x}}{x+1}\right) = \pm\lim _{x\to \infty }\left(\frac{\frac{1}{\sqrt{x}}}{1+\frac{1}{x}}\right)=\pm\frac{0}{1} = 0$$

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@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 $1/{\sqrt x}$.

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