I have to prove that
lim
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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