The Limit Exists!

Since $\int_1^\infty \frac{dx}{(1+x^2)^n} \; < \; 2^{1-n} \int_0^\infty \frac{dx}{x^2+1} \; = \; \frac{\pi}{2^n}$ we deduce that $L \; = \; \lim_{n \to \infty} \sqrt{n}\int_0^1 \frac{dx}{(x^2+1)^n} \; =\; \lim_{n\to\infty}\sqrt{n} \int_0^\infty \frac{dx}{(x^2+1)^n}$ The change of variable $u = x^2$ gives $\int_0^\infty \frac{dx}{(x^2+1)^n} \; = \; \frac12\int_0^\infty \frac{u^{-\frac12}\,du}{(1 + u)^n} \; = \; \tfrac12B(\tfrac12,n-\tfrac12) \; =\; \frac{\sqrt{\pi}\Gamma(n-\frac12)}{\Gamma(n)} \; = \; \frac{\pi}{2^{2n-1}}\binom{2n-2}{n-1}$ and Stirling's Approximation now gives $L \; = \; \lim_{n \to \infty} \frac{\pi \sqrt{n}}{2^{2n-1}}\binom{2n-2}{n -1} \; = \; \tfrac12\sqrt{\pi}$ and so $\tfrac12 < L < 2$ .

Observe that $\displaystyle \left(1+x^{2}\right)^{n}>1+nx^{2}$ . So,

$L=\sqrt{n}\int_{0}^{1}\frac{dx}{\left(1+x^{2}\right)^{n}}<\sqrt{n}\int_{0}^{1}\frac{dx}{1+\left(x\sqrt{n}\right)^{2}}=\arctan\left(\sqrt{n}\right)$

Now, making $n$ tend to $\infty$ we get $\displaystyle L<\frac{\pi}{2}$ hence, the answer being $\displaystyle \frac{1}{2}<L<2$