Elementary question about the limit $\big( 1 - \frac 1 {\sqrt n}\big )^n$, $n\to\infty$. When calculating the limit $L=\big( 1 - \frac 1 {\sqrt n}\big)^n$, $n\to\infty$, what allows me to do the following:
$$
L=\lim \left(\left(1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n}
$$
As the term inside the outer parenthesis goes to $e^{-1}$, we have $L=\lim e^{-\sqrt n}=0$.
It's like we're distributing the parenthesis somehow:
$$\lim \left(\left(1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n}=\lim \left(\lim\left (1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n}=\lim e^{-\sqrt n}$$
The question is: Why can we do this? Which property are we using?

One may use, as $x \to 0$, the classic Taylor expansion:
$$
\ln (1+x)=x-\frac{x^2}2+O(x^3)
$$ giving, as $n \to \infty$,
$$
n\ln \left(1-\frac1{\sqrt{n}}\right)=n \times\left(-\frac1{\sqrt{n}}-\frac1{2n}+O\left(\frac1{n^{3/2}}\right)\right)=-\sqrt{n}-\frac12+O\left(\frac1{\sqrt{n}}\right)
$$ then, as $n \to \infty$,
$$
\left(1-\frac {1}{\sqrt n}\right)^n=e^{-\sqrt{n}-1/2}\left(1+O\left(\frac1{\sqrt{n}}\right)\right) \to 0.
$$
