Evaluate $\space\lim\limits_{n\to\infty}\sqrt{n}\int\limits_{-\infty}^{+\infty}\frac{\cos t}{\left(1+t^2\right)^n}dt$

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

$$ \lim_{n\rightarrow \infty} \sqrt{n}I(n)= \lim_{n\rightarrow \infty} \sqrt{n}\frac{\sqrt{\pi}}{\sqrt{n}}=\sqrt{\pi} $$