Why $\lim_{n\to\infty}(1-\frac{\sigma^2\xi^2}{2n}+o(\frac1n))^n= e^{-\frac{\sigma^2\xi^2}{2}}$

Question

Why is $\lim\limits_{n\to\infty}\displaystyle\Big(1-\frac{\sigma^2\xi^2}{2n}+o(\frac1n)\Big)^n=\large e^{-\frac{\sigma^2\xi^2}{2}}$ ?

Why has $o(\frac{1}{n})$ no effect on the term ? Can I also conclude that $\lim\limits_{n\to\infty}\displaystyle\Big(1+o(\frac1n)\Big)^n=1$ ?

Answer 1

Look at $$ \bigg(1+\frac{1}{n^{1+ \epsilon}}\bigg)^n = \bigg(1+\frac{1}{n^{1 +\epsilon} } \bigg)^{n^{1+ \epsilon -\epsilon}} = \bigg( 1+\frac{1}{n^{1 +\epsilon} } \bigg)^{n^{1+ \epsilon} \cdot n^{-\epsilon}} \leq e^{n^{-\epsilon}} \to_n 1 $$

EDIT: another way: the lower bound is due to Bernoulli inequality. Since $n > 0$ $$ (1+ o(\frac{1}{n})^n \geq (1+n o(\frac{1}{n})) \to_n 1 $$ upper bound: $$ (1+o(\frac{1}{n}))^n \leq e^{n o(\frac{1}{n})} \to_n 1 $$ this is due to $n o(\frac{1}{n}) \to_n 0$.