Limit exists or not? $\lim \limits_{n \to\infty}\ \left[n-\frac{n}{e}\left(1+\frac{1}{n}\right)^n\right] $

Note that we can write

$$\begin{align} n-\frac ne\left(1+\frac1n\right)^n&=n-\frac ne e^{n\log\left(1+\frac1n\right)}\\\\ &=n-\frac ne e^{n\left(\frac1n -\frac{1}{2n^2}+O\left(\frac{1}{n^3}\right)\right)}\\\\ &=n-n\left(1-\frac{1}{2n}+O\left(\frac{1}{n^2}\right)\right) \\\\ &=\frac12+O\left(\frac1n\right) \end{align}$$

This is very similar to Dr. MV's answer using a slightly different approach.

Considering $$A_n=n-\frac{n}{e}\left(1+\frac{1}{n}\right)^n$$ Let us first look at $$B_n=\left(1+\frac{1}{n}\right)^n$$ Take the logarithm $$\log(B_n)=n\log\left(1+\frac{1}{n}\right)$$ Since $n$ is large, use Taylor for $\log(1+x)$ when $x$ is small and replace $x$ by $\frac 1n$. So, you have $$\log(B_n)=n\Big(\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{3 n^3}+O\left(\frac{1}{n^4}\right)\Big)=1-\frac{1}{2 n}+\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ Now $$B_n=e^{\log(B_n)}=e-\frac{e}{2 n}+\frac{11 e}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ Back to $A_n$ $$A_n=n-\frac n e\Big(e-\frac{e}{2 n}+\frac{11 e}{24 n^2}+O\left(\frac{1}{n^3}\right) \Big)=\frac{1}{2}-\frac{11}{24 n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and also how it is approached.

For illustration purposes, using $n=10$, $A_n\approx 0.458155$ while the above formula gives $\frac{109}{240}\approx 0.454167$.