Question: $L$ = $\lim\_{n\to\infty} [n - \frac{n}{e}(1+\frac{1}{n})^n]$

I solved it as follows: 

$L$ = $\lim\_{n\to\infty} [n - \frac{n}{e}(1+\frac{1}{n})^n$]

$L$ = $\lim\_{n\to\infty} [loge^n - loge^{\frac{n}{e}(1+\frac{1}{n})^n}$]

$L$ = $\lim\_{n\to\infty} [loge^n - {\frac{1}{e}}{(1+\frac{1}{n})^{n}}loge^n$]

$L$ = $\lim\_{n\to\infty} loge^n$ - ${\frac{1}{e}}$$\lim\_{n\to\infty}{(1+\frac{1}{n})^{n}}loge^{n}$

$L$ = $\lim\_{n\to\infty} loge^n$ - $\lim\_{n\to\infty}loge^{n}$

$L$ = $0$

But, the answer given is $L = 0.5$. Can anyone please tell me where did I go wrong and help me calculate the right answer?

Answer: 
$\infty-\infty$ is indetermonate expression, one has to find this limit carefully, here is one way:

Let $n=1/t$, then
$$L=\lim\_{t\rightarrow 0} \left(1/t-\frac{1}{et}(1+t)^{1/t}\right),$$

Using the Mclaurin Expansion: $(1+t)^{1/t}=e-et/2+11et^2/24+...$, we get
$$L=1/t-\frac{1}{et}(e-et/2+11et^2/24)= \lim\_{t\rightarrow 0}( 1/2-11t/24+...)=1/2.$$
*Another way to do it (very similar to Dr Zafar Ahmed DSc's answer) .*

Consider
$$a\_n=n - \frac{n}{e}\left(1+\frac{1}{n}\right)^n$$ and let use
$$b\_n=\left(1+\frac{1}{n}\right)^n\implies \log(b\_n)=n \log\left(1+\frac{1}{n}\right)$$
Using Taylor series for $\log(1+\epsilon)$ we then have
$$\log(b\_n)=n\left(\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{3
 n^3}+O\left(\frac{1}{n^4}\right) \right)=1-\frac{1}{2 n}+\frac{1}{3
 n^2}+O\left(\frac{1}{n^3}\right)$$ Continue with Taylor
$$b\_n=e^{\log(b\_n)}=e-\frac{e}{2 n}+\frac{11 e}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ Back to $a\_n$
$$a\_n=n-\frac n e \left(e-\frac{e}{2 n}+\frac{11 e}{24 n^2}+O\left(\frac{1}{n^3}\right) \right)=\frac{1}{2}-\frac{11}{24 n}+O\left(\frac{1}{n^2}\right)$$