Finding $\lim_{n\to \infty}\sqrt n \int_0^1 \frac{\,dx}{(1+x^2)^n}$

No. With the change of variable $ t = \sqrt{n} x $ you get $$ \sqrt{n} \int_0^1 \frac{1}{(1+x^2)^n} \, dx = \int_0^\sqrt{n} \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \, dt = \int_0^{+\infty} \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \chi_{[0,\sqrt{n}]}(t) \, dt. $$Observe that $$ \lim_{n \to \infty} \left( 1 + \frac{t^2}{n} \right)^n \chi_{[0,\sqrt{n}]}(t) = e^{t^2}$$pointwise everywhere, say for $t>0$; also the sequence $ n \mapsto (1 + t^2 /n)^n$ is increasing for all $t \in \mathbb{R}$ which implies that $$\frac{1}{1+t^2} \ge \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \ge \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \chi_{[0,\sqrt{n}]}(t) $$ for all $n \in \mathbb{N}$ and for all $t \in \mathbb{R}$. By the Dominated Convergence Theorem, since $1/(1+t^2) \in L^1([0,+\infty))$, we have that $$\lim_{n \to \infty} \int_0^{+\infty} \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \chi_{[0,\sqrt{n}]}(t) \, dt = \int_0^{+\infty} e^{-t^2} \, dt = \frac{\sqrt{\pi}}{2}. $$

Nope, the limit cannot be zero. In a right neighbourhood of the origin $\frac{1}{1+x^2}\approx e^{-x^2}$, and for large values of $n$ we have that $\int_{0}^{1}e^{-nx^2}\,dx$ is horribly close to $\int_{0}^{+\infty}e^{-nx^2}\,dx$, which scales like $\frac{K}{\sqrt{n}}$ for a positive constant $K$. This actually is the main idea of the Laplace/Hayman methods. In our case

$$ \int_{0}^{1}\frac{dx}{(1+x^2)^n}\stackrel{x\mapsto\tan\theta}{=}\int_{0}^{\pi/4}\cos^{2n-2}(\theta)\,d\theta $$ is at most $\frac{1}{2^{n-1}}\cdot\frac{\pi}{4}$ apart from $$ \int_{0}^{\pi/2}\cos^{2n-2}(\theta)\,d\theta = \frac{\pi}{2\cdot 4^{n-1}}\binom{2n-2}{n-1}=\frac{\pi n}{(2n-1)4^n}\binom{2n}{n}, $$ and since $\frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi n}}$ (by Wallis product or similar elementary manipulations) we have $$ \lim_{n\to +\infty}\sqrt{n}\int_{0}^{1}\frac{dx}{(1+x^2)^n}=\color{red}{\frac{\sqrt{\pi}}{2}}.$$

Let \begin{align*} I_n&=\int_0^\infty\frac{dx}{(1+x^2)^n}=\left.\frac{x}{(1+x^2)^n}\right|_{x=0}^\infty+\int_0^\infty\frac{2nx^2\,dx}{(1+x^2)^{n+1}}\\ &=2n\int_0^\infty\frac{dx}{(1+x^2)^n}-2n\int_0^\infty\frac{dx}{(1+x^2)^{n+1}}=2nI_n-2nI_{n+1} \end{align*} So we get \begin{align*}I_{n+1}&=\frac{2n-1}{2n}I_n=\frac{2n-1}{2n}\frac{2n-3}{2n-2}I_{n-1}=\cdots\\ &=\frac{(2n-1)!!}{(2n)!!}I_1=4^{-n}\binom{2n}{n}\frac \pi 2 \end{align*} By Stirling's formula, $$\int_0^1\frac{dx}{(1+x^2)^{n+1}}=I_{n+1}+O(\tfrac 1 n)\sim \frac{1}{\sqrt{\pi n}}\cdot\frac\pi 2=\frac{\sqrt\pi}{2\sqrt n}$$ Thus the original limit equals $$\sqrt n\int_0^1\frac{dx}{(1+x^2)^n}\sim\frac{\sqrt{\pi n}}{2\sqrt{n-1}}\to\frac{\sqrt \pi}{2}=0.8862\ldots$$