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
$$L=1/t-\frac{1}{et}(e-et/2+11et^2/24)= \lim\_{t\rightarrow 0}( 1/2-11t/24+...)=1/2.$$
You can also find the limit using L'Hospital:

* Write $n - \frac{n}{e}(1+\frac{1}{n})^n = \frac{1}{e}\frac{e-(1+\frac{1}{n})^n}{\frac{1}{n}} \stackrel{x=\frac{1}{n}}{=}-\frac{1}{e}\color{blue}{\frac{(1+x)^{\frac{1}{x}}-e}{x}}$
* Note, that $f(x) = \begin{cases}(1+x)^{\frac{1}{x}} & x>-1,x \neq 0 \\ e & x=0\end{cases}$ is differentiable at $x=0$, so you need only find $\lim\_{x\to 0}\frac{(1+x)^{\frac{1}{x}}-e}{x}$.

\begin{eqnarray\*} \frac{(1+x)^{\frac{1}{x}}-e}{x}
& \stackrel{L'Hosp}{\sim} & \frac{((1+x)^{\frac{1}{x}})'}{1}\\
& = & \underbrace{\frac{(1+x)^{\frac{1}{x}}}{x+1}}\_{\stackrel{x\to 0}{\rightarrow}e}\frac{x-(1+x)\ln(1+x)}{x^2} \\
& \stackrel{L'Hosp}{\sim} & e\frac{1-(\ln(1+x)+1)}{2x} =-e\frac{\ln(1+x)}{2x}\\
& \stackrel{L'Hosp}{\sim} & -e\frac{1}{2(1+x)} \stackrel{x \to 0}{\longrightarrow}\boxed{\color{blue}{-\frac{e}{2}}}
\end{eqnarray\*}
Putting all together:
$$\lim\_{n\to\infty} [n - \frac{n}{e}(1+\frac{1}{n})^n] = -\frac{1}{e}\left(\color{blue}{-\frac{e}{2}}\right) = \boxed{\frac{1}{2}}$$