Estimation: An integral from MIT Integration bee 2022 (QF) We need to determine the value of:
$$\lim_{n\to\infty}\sqrt n\cdot \int_{-1/2}^{1/2} (1-3x^2+x^4)^n\mathrm{d}x$$
Since a limit has been given, one suspects that, as usual, the integral is not easy to evaluate in general and some estimates have to be made. Some promising starts (and observations) include:

*

*The polynomial always lies between $5/16$ and $1$ in the given range. Perhaps we can ignore some higher order terms (even $x$ has modulus less than one)? But the problem is that exponent $n$.

*Reduction formula and integration by parts can give a recursion. But all reasonable ways of splitting the integrand seem to complicate matters rather than simplify them.

*We can now take some hints from the limits. Perhaps some trigonometric substitution can aid. A reasonable one is $x^2=\frac 14\sin \theta$ or $x=\frac 12 \cos\theta$ but there is not much one can do due to the quadratic polynomial.

*Factorize the polynomial as $x^2-3x+1$ has all real roots? But I still get stuck.

But I am unable to make any of these work. Does anyone have any suggestions?
 A: Use the substitution $x\leftrightarrow \sqrt{n}x$
$$\lim_{n\to\infty}\int_{-\frac{\sqrt{n}}{2}}^{\frac{\sqrt{n}}{2}} \left(1 - \frac{3x^2}{n}+\frac{x^4}{n^2}\right)^ndx \longrightarrow \int_{-\infty}^\infty e^{-3x^2}\:dx = \sqrt{\frac{\pi}{3}}$$
by dominated convergence.
A: Change variables to $u = \sqrt{n} x$.  Then you have
$$\sqrt{n} \int_{-\sqrt{n}/2}^{\sqrt{n}/2} \left( 1 - {3u^2 \over n} + {u^4 \over n^2} \right)^n {1 \over \sqrt{n}} \: du$$
The $\sqrt{n}$ cancels, leaving
$$\int_{-\sqrt{n}/2}^{\sqrt{n}/2} \left( 1 - {3u^2 \over n} + {u^4 \over n^2} \right)^n \: du$$
Now as $n \to \infty$ the integrand approaches $\exp(-3u^2)$ so you can use the Gaussian integral $\int_{-\infty}^\infty e^{-x^2} \: dx = \sqrt{\pi}$ to get the answer.
This is an example of Laplace's method for approximating integrals.
A: The other two answers have illustrated very nicely how the Dominated Convergence Theorem can be used to simplify very elegantly the limit in question.
However, both answers are incomplete as they do not show that application of the Dominated Convergence Theorem is justified in this case, i.e. they have not shown that the supremum of the absolute values of the integrands is integrable. This incompleteness can lead to confusion for readers that are not familiar with the DCT.
One might leave the proof that the supremum of the integrands is in $L^1$ as an exercise to the reader (which was unfortunately not done in either of the answers). Since there are some annoying calculations to be made, my answer will solve this exercise by showing that application of the DCT (and also the monotone convergence Theorem!) is justified.

Since
$$\int_{-\frac{\sqrt{n}}{2}}^{\frac{\sqrt{n}}{2}} \left(1 - \frac{3x^2}{n}+\frac{x^4}{n^2}\right)^n\,\mathrm dx = 2\int_0^{\frac{\sqrt{n}}{2}}  \left(1 - \frac{3x^2}{n}+\frac{x^4}{n^2}\right)^n\,\mathrm dx,$$ the integrands in question are
\begin{equation*}\begin{split}f_n:[0,\infty)&\to\mathbb R_+,\\x&\mapsto \left(1 - \frac{3x^2}{n}+\frac{x^4}{n^2}\right)^n [x\le \sqrt n/2],\end{split}\end{equation*}
where I am using Iverson brackets for my notation. I claim that the $f_n$ are an increasing sequence of functions, so that they are dominated by $x\mapsto \exp(-3x^2)$ (and alternatively one could also apply the monotone convergence Theorem without needing to invoke the DCT).
Fix $x\in(0,\infty)$. We calculate
$$\frac{\partial}{\partial n}f_n(x) = \left(\frac{x^4}{n^2}-\frac{3 x^2}{n}+1\right)^n \left(\ln \left(\frac{x^4}{n^2}-\frac{3
   x^2}{n}+1\right)+\frac{n \left(\frac{3 x^2}{n^2}-\frac{2 x^4}{n^3}\right)}{\frac{x^4}{n^2}-\frac{3
   x^2}{n}+1}\right).$$
I claim that $\frac{\partial}{\partial n}f_n(x)\ge 0$ when $x\in(0,\sqrt n/2)$, for which it is enough to show that
$$\ln \left(\frac{x^4}{n^2}-\frac{3
   x^2}{n}+1\right)+\frac{n \left(\frac{3 x^2}{n^2}-\frac{2 x^4}{n^3}\right)}{\frac{x^4}{n^2}-\frac{3
   x^2}{n}+1}\ge 0.$$
Set $y = \frac{x}{\sqrt n}$, then we need to show that
$$g(y)\overset{\text{Def.}}=\frac{3 y^2-2 y^4}{y^4-3 y^2+1}+\log \left(y^4-3 y^2+1\right)\ge 0$$ for all $y\in[0,1/2]$.
But $$g'(y) = \frac{2 y^3 \left(2 y^4-6 y^2+7\right)}{\left(y^4-3 y^2+1\right)^2}$$ is $\ge 0$ for all $y\in(0,1/2)$, therefore we infer from $g(0)=0$ that $g(y)\ge 0$ for all $y\in[0,1/2]$.
It follows that $$\frac{\partial}{\partial n}f_n(x)\ge 0$$ for all $x\in(0,\sqrt n/2)$.
Since $n\mapsto [0\le x\le \sqrt n/2]$ is increasing for all fixed $x\in\mathbb R_+$, it follows that $n\mapsto f_n(x)$ is increasing for all $x\in\mathbb R_+$. Thus we can apply DCT or the monotone convergence Theorem, as claimed.
