Limit $\lim\limits_{n\to\infty}\left(1+\frac{1}{n}+a_n\right)^n=e$ if $\lim\limits_{n\to\infty}na_n=0$

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 $\ln(1+x)$

$$ \ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\dots. $$

$$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}+a_n\right)^n$$

$$= \left(\lim_{n\rightarrow\infty}\left(1+\frac{1+n\cdot a_n}n\right)^{\frac n{1+n\cdot a_n}}\right)^{\lim_{n\rightarrow\infty}(1+n\cdot a_n)}$$

$$= \left(\lim_{y\rightarrow\infty}\left(1+\frac1y\right)^y\right)^{\lim_{n\rightarrow\infty}(1+n\cdot a_n)}$$

$$=e^1$$

as $\lim_{n\rightarrow\infty} n\cdot a_n=0,n\rightarrow\infty \implies y=\frac n{1+n\cdot a_n}\to\infty$