Uniform convergence of ${(1-\frac{x^2}{n})^n}$

Using the basic inequalities $e^y \geqslant 1 +y$ and $e^{-y} \geqslant 1 - y$ for $0 \leqslant y \leqslant 1$ we can show that for $x^2 < n$,

$$\left(1- \frac{x^2}{n}\right)^n \leqslant e^{-x^2}\leqslant \left(1+ \frac{x^2}{n}\right)^{-n}.$$

Applying the Bernoulli inequality $(1 - x^4/n^2)^n \geqslant 1 - x^4/n,$ we get

$$0 \leqslant e^{-x^2} - \left(1- \frac{x^2}{n}\right)^n = e^{-x^2}\left[1 - e^{x^2}\left(1- \frac{x^2}{n}\right)^{n}\right]\\ \leqslant e^{-x^2}\left[1 - \left(1+ \frac{x^2}{n}\right)^{n}\left(1- \frac{x^2}{n}\right)^{n}\right]\\= e^{-x^2}\left[1 - \left(1- \frac{x^4}{n^2}\right)^{n}\right]\leqslant e^{-x^2}\frac{x^4}{n}.$$

For $x \in [-L,L]$, the RHS is bounded by $L^4/n$, which proves uniform convergence.