How can I find $\lim_{n\to\infty}\int_0^\infty\frac{n\cos^2(x/n)}{n+x^4}dx$?

Substituting $x:=n^{1/4} t$ $\>(0<t<\infty)$ gives $$\int_0^\infty{n\cos^2(x/n)\over n+x^4}\>dx=n^{1/4}\int_0^\infty {\cos^2(tn^{-3/4})\over 1+t^4}\ dt\ .\tag{1}$$ The function $t\mapsto \cos^2(tn^{-3/4})$ is nonnegative and $\geq{1\over2}$ for $0\leq t\leq{\pi\over4}n^{3/4}$. It follows that the integral on the right hand side of $(1)$ is $$\geq{1\over2}\int_0^{\pi/4}{dt\over1+t^4}\>dt=:C>0$$ for all $n\geq1$. This proves that the limit in question is $=\infty$.

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 $(1)$.