Evaluate $\space\lim\limits_{n\to\infty}\sqrt{n}\int\limits_{-\infty}^{+\infty}\frac{\cos t}{\left(1+t^2\right)^n}dt$ Find the limit $$ \large
\space\lim\limits_{n\to\infty}\sqrt{n}\int\limits_{-\infty}^{+\infty}\frac{\cos t}{\left(1+t^2\right)^n}dt
$$  
 A: For big $n$ the integrand is cleary dominated from the region around $t=0$. we therefore might write ($\epsilon\ll 1$ )
$$
I(n)\approx \int_{-\epsilon}^{\epsilon}\cos(t)\frac{1}{(1+t^2)^n}=\int_{-\epsilon}^{\epsilon}\cos(t)e^{-n\log(1+t^2)}\approx\int_{-\epsilon}^{\epsilon}\cos(t)e^{-nt^2}\approx\int_{-\epsilon}^{\epsilon}e^{-nt^2}
$$
where we have used $\log(1+x)\approx x$ and $\cos(x)\approx 1$ for $x\ll1$.
The usual trick is now that we can extend the limits of integration back to $\pm \infty$ inducing only an exponentially small error (all steps can be made rigourous by the method of steepest descent). We are therefore left with a standard Gaussian integral
$$
I(n)\approx\int_{-\infty}^{\infty}e^{-nt^2}=\frac{\sqrt{\pi}}{\sqrt{n}}
$$
and we can conclude that
$$
\lim_{n\rightarrow \infty} \sqrt{n}I(n)= \lim_{n\rightarrow \infty} \sqrt{n}\frac{\sqrt{\pi}}{\sqrt{n}}=\sqrt{\pi}
$$
A: Or you could just use $x=t\sqrt{n}$ followed by the dominated convergence theorem to get$$\lim_{n\to\infty}\int_{\Bbb R}(1+\tfrac{x^{2}}{n})^{-n}\cos\tfrac{x}{\sqrt{n}}dx=\int_{\Bbb R}e^{-x^2}dx=\sqrt{\pi}.$$
