$\lim_{n \to \infty} \frac{\int_{\epsilon}^{1}(f(x))^{\frac{n-1}{2}}dx}{\int_{0}^{1}(f(x))^{\frac{n-1}{2}}dx} = 0$ I wish to show that $\lim_{n \to \infty} \frac{\int_{\epsilon}^{1}(1-t^2)^{\frac{n-1}{2}}dt}{\int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt} = 0$ where $0 < \epsilon < 1$.
I've failed to see a simple proof and would appreciate any help.
This isn't the original question, but my efforts led me to this.
 A: We can observe that
$$
I_n(\epsilon)=\sqrt{n}\int_\epsilon^1 (1-t^2)^{\frac{n}{2}}dt = \int_{\sqrt{n}\epsilon}^\sqrt{n} (1-\frac{u^2}{n})^{\frac{n}{2}}du=\int_0^\infty (1-\frac{u^2}{n})^{\frac{n}{2}}1_{(\sqrt{n}\epsilon,\sqrt{n})}(u)du,
$$ for $\epsilon\in (0,1]$. Note that the integrand is dominated by
$$
0\le  (1-\frac{u^2}{n})^{\frac{n}{2}}1_{(\sqrt{n}\epsilon,\sqrt{n})}(u)\leq e^{-\frac{u^2}{2}}.
$$ Since $(1-\frac{u^2}{n})^{\frac{n}{2}}1_{(\sqrt{n}\epsilon,\sqrt{n})}(u)\to 0,\;\forall u\ge 0$ as $n\to\infty$ and $u\mapsto e^{-\frac{u^2}{2}}$ is integrable, by Lebesgue's dominated convergence theorem, we have
$$
\lim_{n\to\infty}I_n(\epsilon)=0,\quad\forall \epsilon>0.
$$
On the other hand, by the same argument, we have
$$
\lim_{n\to\infty}I_n(0)=\int_0^\infty e^{-\frac{u^2}{2}}du=\sqrt{\frac{\pi}{2}}>0.
$$ This implies that
$$
\lim_{n \to \infty} \frac{\int_{\epsilon}^{1}(1-t^2)^{\frac{n-1}{2}}dt}{\int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt}=\lim_{n \to \infty}\frac{I_{n-1}(\epsilon)}{I_{n-1}(0)}=0,
$$ for all $\epsilon>0$.
