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I would like to prove the following limit $$\lim_{n\rightarrow +\infty}\int_0^1(\cos\sqrt x)^n\,dx=0.$$

I know that I can prove it simply using the dominated convergence theorem, but I want to see if there are other ways, involving less sophisticated tools (theorems from basic calculus).

Can someone help me? Any hint would be highly appreciated! Thanks!

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Yet another way: \begin{align} 0&\le\int_0^1\cos^n\sqrt{x}\,dx=[t=\sqrt{x}]=2\int_0^1t\cos^nt\,dt=2\int_0^1\frac{t}{\sin t}\cdot\cos^nt\sin t\,dt\le\\ &\le\frac{2}{\sin 1}\int_0^1\cos^nt\sin t\,dt=\frac{2}{\sin 1}\left[-\frac{\cos^{n+1}t}{n+1}\right]_0^1=O\left(\frac{1}{n+1}\right)\to 0. \end{align}

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Try to show, for example, that $0\leq \cos(\sqrt{x})\leq 1-x/4$ on the interval. Then just integrate and squeeze.

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    $\begingroup$ Another estimation works nicely without binomial: $\cos t\le\exp(-t^2/2)$, $$ \int_0^1\cos^n\sqrt{x}\,dx\le\int_0^1e^{-nx/2}\,dx\le\int_0^\infty e^{-nx/2}\,dx=\frac{2}{n}. $$ $\endgroup$ – A.Γ. Jan 30 '18 at 17:38
  • $\begingroup$ @A.Γ. Thank you my friend. That is indeed more elegant. Frankly, I just took the 1/4 arbitrarily. I was a bit upset. Not with OP, but with his/her education. These kind of estimates should be very straight forward and natural, before one even meet anything like dominated convergence. $\endgroup$ – mickep Jan 30 '18 at 17:46
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Pick $0 <a <1$. Then $$\int_0^1(\cos\sqrt x)^n\,dx\leq \int_0^a(\cos\sqrt x)^n\,dx+\int_a^1(\cos\sqrt x)^n\,dx\leq a+(1-a) \cos(\sqrt{a})^n$$

Now set $n \to \infty$ to get $$\limsup_{n \to \infty} \int_0^1(\cos\sqrt x)^n\,dx\leq a$$ (you need to use limsup because you don't know that the limit exists).

Since this is true for all $a >0$, you can conclude that $$\limsup_{n \to \infty} \int_0^1(\cos\sqrt x)^n\,dx=0$$ from which the claim follows immediately.

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  • $\begingroup$ This is the most straightforward way to understand what happens here. $\endgroup$ – A.Γ. Jan 30 '18 at 17:51

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