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

Question

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

My Attempt : Now $(1-x^2)^n$ is uniformly convergent to $0$ when $x \in (0 , \delta)$. So $\lim_{n\to \infty}{\int_{\delta}^1 (1-x^2)^n} = 0$. Here $0 < \delta < 1$. IS it sufficient to prove $\lim_{n\to \infty}{\int_{0}^1 (1-x^2)^n} = 0$ ?

Can anyone please help me ??

Answer 1

$$I_n=\int_0^1(1-x^2)^ndx$$ $u=1-x^2,du=-2xdx=-2\sqrt{1-u}dx\Rightarrow dx=-2\frac{du}{\sqrt{1-u}}$ and so we have: $$I_n=2\int_0^1\frac{u^n}{\sqrt{1-u}}du$$ now on the lower part we have an issue as $u\to1$ and also on the top $\to1$ but for the rest of the interval it is clear that: $$\lim_{n\to\infty}\frac{u^n}{\sqrt{1-u}}\to0\,\,\,\,u\ne1$$

Answer 2

Put $Z_n = \int_0^n{(1-x^2)^n dx}$. We have $\forall \epsilon >0$ $$Z_n= \int_0^n{(1-x^2)^n dx} = \int_0^{\frac{\epsilon}{2}}{(1-x^2)^n dx}+\int_{\frac{\epsilon}{2}}^{1}{(1-x^2)^n dx} \leq \frac{\epsilon}{2} + \int_{\frac{\epsilon}{2}}^{1}{(1-(\frac{\epsilon}{2})^2)^n dx} \leq \frac{\epsilon}{2} + (1-\frac{\epsilon^2}{4})^n $$

And we have $\forall n> \frac{\ln (\frac{\epsilon}{2})}{\ln (1-\frac{\epsilon^2}{4})}$, $(1-\frac{\epsilon^2}{4})^n \leq \frac{\epsilon}{2}$.

So, $\forall \epsilon >0, \exists N= \frac{\ln (\frac{\epsilon}{2})}{\ln (1-\frac{\epsilon^2}{4})}$ so that $\forall n>N , Z_n < \frac{\epsilon}{2} +\frac{\epsilon}{2} =\epsilon $. We can conclude $Z_n \longrightarrow 0 $