Q:

$\lim_{n \to \infty} \frac{\int_{\epsilon}^{1}(f(x))^{\frac{n-1}{2}}dx}{\int_{0}^{1}(f(x))^{\frac{n-1}{2}}dx} = 0$

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:

First of all, we will explicitly compute $\int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt$ using a dirty trick. 
Defining $ I_n: =\int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt$ and integrating by parts
\begin{align*}I_n &=  \int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt\\
&=\left.t(1-t^2)^{\frac{n-1}{2}}\right|_{0}^{1}+(n-1)\int_{0}^{1}t^2(1-t^2)^{\frac{n-1}{2}-1}dt\\
&=0+(n-1)\int_{0}^{1}(1 - ( 1-t^2))(1-t^2)^{\frac{n-1}{2}-1}dt\\
&=(n-1) \int_{0}^{1}(1-t^2)^{\frac{(n-2)-1}{2}} + (n-1) \int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt\\
&= (n-1)I_{n-2} + (n-1)I_n,
\end{align*}
which (miraculously) implies 
$$I_n = \frac{n-1}{n} I_{n-2}. $$
Since $I_1 = 1$ and $I_2 = \frac{\pi}{4}$, we have achieved 
$$\displaystyle\int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt = 
\left\{\begin{array}{l}
\displaystyle\frac{2}{3}\cdot \frac{4}{5}\cdot ... \cdot \frac{2k}{2k+1},\ \mbox{if} \ n=2k+1,\\
\displaystyle\frac{\pi}{4} \cdot \frac{3}{4}\cdot \frac{5}{6} \cdot ...\cdot \frac{2k-1}{2k},\  \mbox{if} \ n=2k.
\end{array}\right.$$
On the other hand
$$ \int_{\varepsilon}^{1}(1-t^2)^{\frac{n-1}{2}}dt \leq (1-\varepsilon) (1-\varepsilon)^{\frac{n-1}{2}} =(1-\varepsilon)^{\frac{n+1}{2}}, $$
then
$$\frac{\int_{\epsilon}^{1}(1-t^2)^{\frac{n-1}{2}}dt}{\int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt} \leq \frac{(1-\varepsilon)^{\frac{n+1}{2}}}{\int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt}. $$
Joining all the previous results, to prove your statement we just need to show that the below limits are $0$ (define $\alpha = 1-\varepsilon$ just to simplify 
 the notation and remember that $0<\alpha<1$).
$$(1) \quad \lim_{k\to \infty} \frac{\alpha^{\frac{2k+1}{2}}}{\int_{0}^{1}(1-t^2)^{\frac{2k-1}{2}}dt} =  \lim_{k\to \infty} \alpha^{\frac{2k+1}{2}} \cdot \frac{4}{\pi} \cdot \frac {4} {3}\cdot \frac{6}{5} \cdot ...\cdot \frac {2k} {2k-1} .  $$
$$(2)\quad   \quad \lim_{k\to \infty} \frac{\alpha^{\frac{(2k+1)+1}{2}}}{\int_{0}^{1}(1-t^2)^{\frac{(2k+1)-1}{2}}dt} =  \lim_{k\to \infty} \alpha^{k+1} \cdot \frac{3}{2}\cdot \frac{5}{4}\cdot ... \cdot \frac{2k+1}{2k}.$$

Computing the limit of (1)
We will show that the limit of (1) is equal to $0$. Since $$\lim_{k\to \infty} \frac{2k}{2k-1} =1,$$
there exists $n_0 \in \mathbb{N}$ such that $$\frac{2n}{2n+1}\cdot \alpha < \frac{1+\alpha}{2} <1, \forall \ n > n_0.$$
Defining the constant $$C = \alpha^{n_0} \cdot  \frac{4}{3}\cdot \frac{6}{5} \cdot ...\cdot \frac{2n_0}{2n_0-1}, $$ 
we achieve the useful inequality
\begin{align*}
0\leq\alpha^{k + n_0+1} \cdot \frac{4}{3}\cdot \frac{6}{5} \cdot ...\cdot \frac{2(k+n_0)}{2(k+n_0)-1} &= \left(\prod_{n=1}^{n_0} \alpha \cdot \frac{2n}{2n-1}\right) \cdot \left(\prod_{k=n_0}^{2(n_0+k)+1} \alpha \cdot \frac{2k}{2k-1}\right)\\
&\leq C \cdot \left(\frac{1+\alpha}{2}\right)^k
\end{align*}
Then, if $k\rightarrow \infty$, we conclude that $$\lim_{k\to \infty} \alpha^{k+1} \cdot  \frac{4}{3}\cdot \frac{6}{5} \cdot ...\cdot \frac{2k}{2k-1} =0.$$
Finally
\begin{align*}
\lim_{k\to \infty} \alpha^{\frac{2k+1}{2}} \cdot \frac{4}{\pi} \cdot \frac {4} {3}\cdot  ...\cdot \frac {2k} {2k-1} =\frac{4}{\alpha^{1/2}\pi} \cdot \lim_{k\to \infty} \alpha^{k+1} \cdot \frac {4} {3}\cdot...\cdot \frac {2k} {2k-1}=0
\end{align*}
using previous results.

Computing the limit of (2)
We will show that the limit of (2) is equal to $0$. Since $$\lim_{k\to \infty} \frac{2k+1}{2} =1,$$
there exists $n_1 \in \mathbb{N}$ such that $$\frac{2n+1}{2n}\cdot \alpha < \frac{1+\alpha}{2} <1, \forall \ n > n_1.$$
Defining the constant $$K = \alpha^{n_1} \cdot  \frac{3}{2}\cdot \frac{5}{4} \cdot ...\cdot \frac{2n_1+1}{2n_1}, $$ 
it lead us to the inequality
\begin{align*}
0\leq\alpha^{k + n_1+1} \cdot  \frac{3}{2}\cdot \frac{5}{4} \cdot ...\cdot \frac{2(k+n_1)+1}{2(k+n_1)-1} &= \left(\prod_{n=1}^{n_1} \alpha \cdot \frac{2n+1}{2n}\right) \cdot \left(\prod_{n=n_0}^{2(n_0+k)+1} \alpha \cdot \frac{2n+1}{2k}\right)\\
&\leq K \cdot \left(\frac{1+\alpha}{2}\right)^k.
\end{align*}
Then, if $k\rightarrow \infty$, we conclude that $$\lim_{k\to \infty} \alpha^{k+1} \cdot \frac{3}{2}\cdot \frac{5}{4} \cdot ...\cdot \frac{2k+1}{2k} =0.$$
Since the limits of $(1)$ and $(2)$ are $0$, we proved 
$$\lim_{n\to \infty}\frac{\int_{\epsilon}^{1}(1-t^2)^{\frac{n-1}{2}}dt}{\int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt}=0.$$ 
I hope this answer helps you.

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 $\epsilon>0$.

