We know that $\lim_{n \to \infty}$ $(1+\frac xn)^n=e^x$. How to prove that $\lim_{n \to \infty}$ $(1+\frac xn-o(\frac 1n))^n=e^x$?
Attempt of the proof:
Let $\epsilon>0$ $\exists n_0$ such that $\forall n\ge n_0$ $|o(\frac 1n)|<\epsilon\frac 1n$ and $|n*o(\frac 1n)|<\epsilon$.
Thus $(1+\frac{x+\epsilon}{n})^n>(1+\frac xn-o(\frac 1n))^n>(1+\frac{x-\epsilon}{n})^n$.
So $e^x*e^\epsilon \ge \lim_{n \to \infty}$ $(1+\frac xn-o(\frac 1n))^n \ge \frac{e^x}{e^\epsilon}$ $\forall \epsilon>0$.
Say $\lim_{n \to \infty}$ $(1+\frac xn-o(\frac 1n))^n=y$, it can not be that $y<e^x$, because if it is so, then $\exists \epsilon>0$ such that $e^\epsilon y<e^x$ and the second inequality above is violated. By the similar talking $y>e^x$ is impossible.
This completes the proof.
Let $a_n = o(\frac{1}{n})$. Let $0 < \varepsilon < e^x$. Choose an $r > 0$ such that $e^{x - r} > e^x - \frac{\varepsilon}{2}$ and $e^{x + r} < e^x + \frac{\varepsilon}{2}$. Since $\lim_{n\to \infty} na_n = 0$, there exists an $N_1\in \Bbb N$ such that $|na_n| < r$ for all $n\ge N_1$. Since $\lim_{n\to \infty} (1 + \frac{x - r}{n})^n = e^{x - r}$ and $\lim_{n\to \infty} (1 + \frac{x + r}{n})^n = e^{x + r}$, there exist $N_2, N_3\in \Bbb N$ such that $(1 + \frac{x-r}{n})^n > e^{x - r} - \frac{\varepsilon}{2}$ for all $n\ge N_3$ and $(1 + \frac{x + r}{n})^n < e^{x + r} + \frac{\varepsilon}{2}$ for all $n\ge N_2$. Let $N = \max\{N_1,N_2,N_3\}$. If $n\ge N$, then $$\left(1 + \frac{x}{n} - a_n\right)^n = \left(1 + \frac{x - na_n}{n}\right)^n > \left(1 + \frac{x - r}{n}\right)^n > e^{x - r} - \frac{\varepsilon}{2} > e^x - \varepsilon$$ and $$\left(1 + \frac{x}{n} - a_n\right)^n = \left(1 + \frac{x - na_n}{n}\right)^n < \left(1 + \frac{x + r}{n}\right)^n < e^{x + r} + \frac{\varepsilon}{2} < e^x + \varepsilon.$$ Thus $$\left\lvert \left(1 + \frac{x}{n} - a_n \right)^n - e^x\right\rvert < \varepsilon \quad \text{if $n\ge N$.}$$ Therefore $$\lim_{n\to \infty} \left(1 + \frac{x}{n} - o\left(\frac{1}{n}\right)\right)^n = e^x.$$