Integral from MIT Integration Bee 2023 Quarterfinals - $\lim_{n\to\infty} \frac{1}{n} \int_{0}^{n} \cos^2(\pi x^2/\sqrt{2}) \, \mathrm{d}x$ This question is from the MIT Integration Bee 2023 Quarterfinal #2. I'm doing these integrals to improve my quick math skills, and was wondering if there's a way to solve this within two minutes.
The goal is to show that $$\lim_{n\to\infty} \frac{1}{n} \int_{0}^{n} \cos^2\left(\frac{\pi x^2}{\sqrt{2}}\right) \,\textrm{d}x = \frac{1}{2} $$
I've tried L'Hospital's Rule then FTC1 but I don't think the first part would even apply here. I'm not sure how to approach such a problem, especially if it's to be solved within two minutes.

Transforming the integral by cosine half-angle formula yields
$$
\begin{aligned}
I(n) :& =\int_0^n \cos ^2\left(\frac{\pi x^2}{\sqrt{2}}\right) d x \\
& =\frac{1}{2} \int_0^n\left[1+\cos \left(\pi \sqrt{2} x^2\right)\right] d x \\
& =\frac{n}{2}+\frac{1}{2} \int_0^n \cos \left(\pi \sqrt{2} x^2\right) d x
\end{aligned}
$$Letting $\sqrt{\pi \sqrt{2}} x \rightarrow x$, then
$$
\begin{aligned}
I(n)&=\frac{n}{2} +\frac{1}{2 \sqrt{\pi \sqrt{2}}} \int_0^{n \sqrt{\pi \sqrt{2}}} \cos \left(x^2\right) d x  \\
\lim _{n \rightarrow \infty}\left[\frac{1}{n} I(n)\right] & =\frac{1}{2}+\frac{1}{2 \sqrt{\pi \sqrt{2}}}\left(\lim _{n \rightarrow \infty} \frac{1}{n}\right)\left(\lim _{n \rightarrow \infty} \int_0^{n \sqrt{\pi \sqrt{2}}} \cos \left(x^2\right) d x\right)\\
& =\frac{1}{2}+\frac{1}{2 \sqrt{\pi \sqrt{2}}} \cdot 0 \cdot \sqrt{\frac{\pi}{8}}\quad  (*) \\
& =\frac{1}{2}
\end{aligned}
$$
where (*) comes from the post.
