<s>[INST] Calculate the following Limit $\lim_{n\rightarrow \infty} \sqrt{n}\int_{0}^{\infty} \frac{1-\cos(x)}{1+nx^6}dx$ Calculate the Limit
$$\lim_{n\rightarrow \infty} \sqrt{n}\int_{0}^{\infty} \frac{1-\cos(x)}{1+nx^6}dx$$
I've been working on trying to find the limit using Dominated Convergence.
If we let $f_n(x) = \frac{\sqrt{n}(1-cos(x))}{1+nx^6} $
then
$$f_n(x)\leq \frac{2\sqrt{n}}{1+nx^6}=\frac{2}{\sqrt{n}^{-1}+\sqrt{n}x^6} $$
but of course we run into trouble around the origin.
Any help would be appreciated.
 [/INST] Hint. By using
$$
1-\cos x=2\sin^2\frac x2
$$ and the change of variable
$$
x=n^{-1/6}u,\quad u^6=nx^6,\quad dx =n^{-1/6}du,
$$one gets
$$
\lim_{n\rightarrow \infty} \sqrt{n}\int_{0}^{\infty} \frac{1-\cos(x)}{1+nx^6}dx=\lim_{n\rightarrow \infty} 2n^{1/3}\int_{0}^{\infty} \frac{\sin^2\left(\frac{u}{2n^{1/6}}\right)}{1+u^6}du
$$then using
$$
\left|2n^{1/3}\sin^2\left(\frac{u}{2n^{1/6}}\right)\right|\le \frac{u^2}{2},\qquad \lim_{n\rightarrow \infty}2n^{1/3}\sin^2\left(\frac{u}{2n^{1/6}}\right)=\frac{u^2}{2},
$$ yields
$$
\lim_{n\rightarrow \infty} \sqrt{n}\int_{0}^{\infty} \frac{1-\cos(x)}{1+nx^6}dx=\frac12\int_{0}^{\infty} \frac{u^2}{1+u^6}du=\frac \pi{12}.
$$
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