Why does $\lim_{n\to \infty}\sqrt{n}(1-x^2)^n=0$ if $0<|x|\leq 1$? I was working on constructing an approximation of the identity, and one point in my construction requires me to show that there is a sufficiently large $n$ such that given $\epsilon>0$ and some $0<\delta<1$, then $\sqrt{n}(1-x^2)^n<\epsilon$ if $\delta\leq|x|\leq 1$.
I know that $\lim_{n\to\infty}\sqrt{n}(1-x^2)^n=0$ since $(1-x^2)$ is a positive number less than $1$, and this goes to $0$ much faster than $\sqrt{n}$ goes to $\infty$. So such a large $n$ does exist.
I'm happy to leave it at that, but how could you more rigorously convince someone that the limit does in fact go to $0$?
 A: Deal with $|x|=1$ separately, it is very easy. Now suppose that $|x|\lt 1$. 
Let $1-x^2=\dfrac{1}{1+a}$. Note that since $0\lt 1-x^2\lt 1$,  $a$ is positive. By the Binomial Theorem, or the easily proved (by induction) Bernoulli Inequality, we have $(1+a)^n \ge 1+an$.
Thus
$$\sqrt{n} (1-x^2)^n \le \frac{\sqrt{n}}{1+an}.$$
Now for any positive $\epsilon$, it is relatively easy to come up with an $N$ such that
$$\frac{\sqrt{n}}{1+an}\lt \epsilon$$
if $n\gt N$. For example, we can take as $N$ any integer $\ge \left(\dfrac{1}{a\epsilon}\right)^2$. 
A: Taking logs, it suffices to show that
$$\lim_{n \to \infty} 1/2 \log n + n \log (1-x^2) =-\infty.$$
For $\log (1-x^2) \neq 0$, the expression above is asymptotic to its linear term (i.e. the logarithmic term $1/2 \log n$ is dominated by the linear term $n \log(1-x^2)$). Our result then follows because $\log (1-x^2)<0$. 
A: Put $y=1-x^2$, then $\sqrt{n}(1-x^2)^n=(yn^{1/(2n)})^n$. As $\lim_{n\to\infty}n^{1/n}=1$, there is some constant $y<d<1$ such that $yn^{1/(2n)}<d$ when $n$ is sufficiently large. Therefore $(yn^{1/(2n)})^n\le d^n\to0$.
