Prove: $\lim\limits_{n \to \infty} \int_{0}^{\sqrt n}(1-\frac{x^2}{n})^ndx=\int_{0}^{\infty} e^{-x^2}dx$

Although perhaps not the best approach here, I give a proof with bare hands. The idea is to bound the integrand $f(x) = \left( 1- \frac{x^2}{n} \right)^n$, and hence the integral, directly.

• Upper bound. Since $1 - z \leqslant \mathrm e^{-z}$ for all $z$, we have we have $f(x) \leqslant \mathrm e^{-x^2}$ for $0 \leqslant x \leqslant \sqrt{n}$. Therefore, $$ \int_{0}^{\sqrt{n}} f(x) \, dx \leqslant \int_{0}^{\sqrt{n}} \mathrm e^{-x^2} \, dx \lt \int_{0}^{\infty} \mathrm e^{-x^2} \, dx. $$

• Lower bound. Here we need the standard estimate $1 -z \geqslant \mathrm e^{-z - z^2}$ valid for $0 \leqslant z \leqslant \frac14$. (This can be obtained by using Taylor's theorem for $\log (1-z)$ at $z=0$.) Therefore, it follows that for $0 \leqslant x \leqslant \frac12 \sqrt{n}$, $$f(x) \geqslant \exp \left( - x^2 - \frac{x^4}{n} \right) .$$ In particular, for $0 \leqslant x \leqslant n^{1/8} \lt \frac12 \sqrt{n}$ (for large enough $n$), we have $ f(x) \geqslant \mathrm{e}^{-x^2} \mathrm e^{- \frac{1}{\sqrt{n}}} $.

  Therefore, $$ \begin{align*} \int_0^{\sqrt{n}} f(x) \, dx &\geqslant \int_0^{n^{1/8}} f(x) \, dx \\ &\geqslant \int_0^{n^{1/8}} \mathrm{e}^{-x^2} \mathrm e^{- \frac{1}{\sqrt{n}}} \, dx \\ &= \mathrm e^{-\frac{1}{\sqrt{n}}} \cdot \int_0^{n^{1/8}} \mathrm{e}^{-x^2}\, dx \end{align*} $$ Notice that the sequence $\mathrm e^{-\frac{1}{\sqrt{n}}} \cdot \int_0^{n^{1/8}} \mathrm{e}^{-x^2}\, dx$ converges to $\int_0^\infty \mathrm e^{-x^2} \, d x$.

Therefore by the sandwich theorem, we can conclude that $$ \int_0^{\sqrt{n}} f(x) \, dx \to \int_0^\infty \mathrm e^{-x^2} \, dx. $$

I fail to see how the behaviour of the integrals on $[a,b]$ could yield the $[0,+\infty)$ case.

Hint for a solution: consider the functions $f$ and $f_n$ defined on $\mathbb R_+$ by $f(x)=\mathrm e^{-x^2}$ and $$ f_n(x)=\left(1-\frac{x^2}n\right)^n\cdot[x\leqslant\sqrt{n}]. $$ Show that the sequence $(f_n)_{n\geqslant1}$ is increasing and that $f_n\to f$ pointwise. Then, find a theorem in your notes which guarantees that, in this setting, the integral of $f_n$ converges to the integral of $f$.