How can I find $\lim_{n\to\infty}\int_0^\infty\frac{n\cos^2(x/n)}{n+x^4}dx$? I am trying to find the value of this integral:
$\displaystyle{\lim_{n\to\infty}\int_0^\infty\frac{n\cos^2(x/n)}{n+x^4}dx}$.
The integrand tends to 1 as $n$ goes to infinity. So if some convergence theorem holds, the integral would tend to infinity(not Lebesgue dominated convergence theorem, since if there were an integrable function dominating the integrand, there would be a contradiction). So I guess the value of this integral should be infinity, but I am not sure.

By a direct integration, we may obtain more than the value of the desired limit.
We have, as $n$ tends to $+\infty$:

$$
\int_0^{+\infty}n\:\frac{\cos^2(x/n)}{n+x^4}{\rm d}x =\color{#008B8B}{\frac{\pi\sqrt{2}}{4}} \color{#006699}{\:n^{1/4}}+\color{#008B8B}{\frac{\pi\sqrt{2}}{4}}\color{#006699}{\frac{1}{n^{5/4}}}+\color{#008B8B}{\frac{\pi}{3}}\color{#006699}{\frac{1}{n^2}}+\mathcal{O}\left(\color{#006699}{\frac{1}{n^{11/4}}}\right) \tag1
$$ 

Of course the desired limit is $+\infty$. 
One may recall that we have, using inverse Laplace transform, the standard result:
$$
\int_0^{+\infty}\frac{\cos^2(ax)}{b^4+x^4}{\rm d}x =\frac{\pi +e^{-\sqrt{2} a b} \pi  \left(\cos\left(\sqrt{2} a b\right)+\sin\left(\sqrt{2} a b\right)\right)}{4 \sqrt{2} \:b^3}\quad a>0,\,b>0.
$$ 
Then putting $a:=\dfrac 1n$ and $b:=n^{1/4}$, we get
$$
\int_0^{+\infty}\frac{\cos^2(x/n)}{n+x^4}{\rm d}x =\frac{\pi +e^{-\Large \frac{\sqrt{2}}{n^{3/4}}} \pi  \left(\cos\left(\Large \frac{\sqrt{2}}{n^{3/4}}\right)+\sin\left(\Large \frac{\sqrt{2}}{n^{3/4}}\right)\right)}{4 \sqrt{2} \:n^{3/4}}.
$$ Now, as $x$ is near $0$, we apply the standard Taylor series expansions:
$$
\begin{align}
e^{-x} & =1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}+\mathcal{O}\left(x^5\right) \\
\cos x & =1-\frac{x^2}{2!}+\frac{x^4}{4!}+\mathcal{O}\left(x^5\right)  \\
\sin x & =x-\frac{x^3}{3!}+\mathcal{O}\left(x^5\right)  
\end{align}
$$
with $\displaystyle x:=\frac{\sqrt{2}}{n^{3/4}} $ giving $(1)$.
