Question: Evaluate $\lim_{n \to \infty} \int _0^\sqrt n (1-\frac {x^2}n)^ndx$
Thoughts We proved that the inner function uniformly converges to $e^{-x^2}$ in the segment but then calculating the integral made that turns out is something we don't know how to do. Hints?
Hint: Write the integral as: $$\lim_{n\to\infty}\int_{-\infty}^\infty\left(1-\frac{x^2}{n}\right)^n\chi_{[0,\sqrt{n}]}(x)dx$$ where $\chi_A$ is the characteristic function of $A$ (i.e. it is $1$ on $A$ and $0$ everywhere else). Then use the dominated convergence theorem to bring the limit inside the integral.
If the inner function converges uniformly to a function then the integral converges uniformly, also. So you just need the probability integral or the properties of the normal distribution function.
Here $$\int_0^\infty e^{-x^2} dx=\sqrt{\pi} \frac{1}{\sqrt{\pi}}\int_0^\infty e^{ -x^2}dx =\sqrt{\pi}/2.$$
$(1- \dfrac{x^{2}}{n})^{n} \le (e^\frac{-x^{2}}{n})^{n} \le e^{-x^{2}} $ because $1-u \le e^{-u}$
$ e^{-x^{2}} $ is integrable and $f_{n}(x)= (1- \dfrac{x^{2}}{n}) $ converges $f(x)= e^{-x^{2}} $ pointwise . By using Dominated Convergence Theorem. so $$ \lim_{n \rightarrow\infty} \int f_{n} d \mu =\int f d\mu $$ then $$ \lim_{n \rightarrow\infty} \int^{\sqrt{n}}_{0} (1- \dfrac{x^{2}}{n}) d x =\int^{\infty}_{0} e^{-x^{2}} dx $$ so lets say $$ A= \int^{\infty}_{0} e^{-x^{2}} dx $$ then $$ A^{2}= \int^{\infty}_{0} \int^{\infty}_{0} e^{-(x^{2}+y^{2})} dx $$... Ok, I think it is enough , can you continue yourself now ?
