What is the proper way to handle the limit with little-$o$?

Question

I was hoping to show that

$$\left(1-\frac{x}{n}+o\left(\frac{2x}{n}\right)\right)^n \xrightarrow{n\to\infty} e^{-x}$$

which would be just fine without the little-$o$.

Trying binomial formula:

$$\left(1-\frac{x}{n}+o\left(\frac{2x}{n}\right)\right)^n = \sum_{k=0}^n \binom{n}{k} \left(1-\frac{x}{n}\right)^{n-k} \left( o\left( \frac{2x}{n} \right) \right)^k$$

Yet it doesn't look like it leads me somewhere where I could use

$$\frac{o\left( \frac{2x}{n} \right)}{\frac{2x}{n}}\xrightarrow{n\to\infty} 0$$

What is the proper way to handle it?

Edited after the comment by Claude Leibovici.

Answer 1

Note that if $\lim_{n\to\infty}f(n)=0$ and $\lim_{n\to\infty}g(n)=\infty$ then $$ \lim_{n\to\infty}(1+f(n))^{g(n)}=e^{\lim_{n\to\infty}f(n)g(n)}. $$ In your case, you have $f(n)=-\frac xn+o\left(\frac{2x}n\right)$ and $g(n)=n$ which of course satisfies the above condition. Since $$ {\lim_{n\to\infty}f(n)g(n)}={\lim_{n\to\infty}}\left\{-\frac xn+o\left(\frac{2x}n\right)\right\}n=-x+0=-x $$ then your limit equals to $e^{-x}$.