Why is $\lim_{n\to\infty} n(e - (1+\frac{1}{n})^n) = \frac{e}{2}$ I'm having trouble understanding why
$$\lim_{n\to\infty} n(e - (1+\tfrac{1}{n})^n) = \frac{e}{2}$$
Can someone offer me a proof for this?
 A: Write that $(1+\frac{1}{n})^n = e^{n \ln(1+\frac{1}{n}) }$
Then factorise by $e$, and use the taylor expansion of $\ln(1+x)$
A: Recall that, as $x \to 0$, we have
$$
\begin{align}
e^x& =1+x+\mathcal{O}(x^2)\\
\ln (1+x)&=x-\frac {x^2}{2}+\mathcal{O}(x^3)
\end{align}
$$ giving, as $n \to \infty$,
$$
n\ln (1+\frac {1}{n})=n \left(\frac {1}{n}-\frac {1}{2n^2}+\mathcal{O}(\frac {1}{n^3})\right)=1-\frac {1}{2n}+\mathcal{O}(\frac {1}{n^2})
$$ and
$$
\begin{align}
n(e - (1+\tfrac{1}{n})^n)&=n(e - e^{1-\frac {1}{2n}+\mathcal{O}(\frac {1}{n^2})})\\\\
&=ne\left( 1 - e^{-\frac {1}{2n}+\mathcal{O}(\frac {1}{n^2})}\right)\\\\
&=ne\left( 1 - \left(1-\frac {1}{2n}+\mathcal{O}\left(\frac {1}{n^2}\right)\right)\right)\\\\
&=\frac{e}{2}+\mathcal{O}\left(\frac {1}{n}\right).
\end{align}
$$
A: Using: $\ln(1 + \frac{1}{n}) \sim \frac{1}{n} - \frac{1}{2n^2}$
And: $e^{-\frac{1}{2n}} \sim 1 - \frac{1}{2n} $
$$n(e - (1 + \frac{1}{n})^n) = n(e - e^{n\ln(1 + \frac{1}{n})}) = ne(1 - e^{n\ln(1 + \frac{1}{n}) - 1}) \sim ne(1 - e^{1 - \frac{1}{2n} - 1}) = ne(1 - e^{-\frac{1}{2n}}) = ne(1 - (1 - \frac{1}{2n}) = ne(\frac{1}{2n}) = \frac{e}{2}$$
$\therefore$ etc.
A: hint: Put $x = \dfrac{1}{n} \to L = \displaystyle \lim_{x \to 0} \dfrac{e-\left(1+x\right)^{\frac{1}{x}}}{x}=\dfrac{-\frac{d}{dx}\left(1+x\right)^{\frac{1}{x}}|_{x=0}}{1}=....$
