Limit $\lim\limits_{n\to\infty}\left(1+\frac{1}{n}+a_n\right)^n=e$ if $\lim\limits_{n\to\infty}na_n=0$ 
Let $\{a_n\}$ be any sequence of real numbers such that $\lim_{n\rightarrow\infty}na_n=0$. Prove that $$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}+a_n\right)^n=e$$

I thought about using binomial theorem. So $\left(1+\dfrac{1}{n}+a_n\right)^n = \left(1+\dfrac{1}{n}\right)^n + na_n\left(1+\dfrac{1}{n}\right)^{n-1} + \dbinom{n}{2}a_n^2\left(1+\dfrac{1}{n}\right)^{n-2}+\ldots$.
The first term has limit $e$, the second term has limit $0$ (because $na_n$ has limit $0$, and $\left(1+\dfrac{1}{n}\right)^{n-1} = \left(1+\dfrac{1}{n}\right)^{n}\left(1+\dfrac{1}{n}\right)^{-1}$ has limit $e$.) But for the other terms, it seems hard to find the limit.

Here is a technique
$$\left(1+\frac{1}{n}+a_n\right)^n= e^{n\ln\left(1+\frac{1}{n}+a_n\right)}= e^{n\left ( (\frac{1}{n}+a_n) -\frac{1}{2}(\frac{1}{n}+a_n)^2+\dots. \right) }=e^{\left( (1+na_n) -\frac{n}{2}(\frac{1}{n}+a_n)^2+\dots.\right)} $$
$$ = e^{\left( (1+na_n) -\frac{n}{2}\frac{(1+na_n)^2}{n^2}+\dots.\right)} .$$
Now, take the limit as $n\to \infty$ gives the limit $e$.
Note: We used the Taylor series of $\ln(1+x)$
$$ \ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\dots. $$
