Evaluation of $L=\lim_{ n \to \infty } \Big(\frac{n^{n+1}\cdot e}{(n+1)^n}-n\Big)$ Find the value of following limit:
$$L=\lim_{ n \to \infty } \Big(\frac{n^{n+1}\cdot e}{(n+1)^n}-n\Big)$$
How to solve this indeterminate form? After taking do we have apply L-Hopital's rule or is there any better approach?
 A: Hint. One may use some standard Taylor series expansions writing, as $n \to \infty$,
$$
\begin{align}
\frac{n^{n+1}.e}{(n+1)^n}&=n\cdot e\cdot e^{\large -n\log(1+1/n)}
\\\\&=n\cdot e\cdot e^{\large -n\left(\frac1n-\frac1{2n^2}+O\left(\frac1{n^3}\right)\right)}
\\\\&=n\cdot e\cdot e^{\large -1+1/(2n)+O(1/n^2)}
\\\\&=n\left(1+\frac{1}{2n}+O\left(\frac{1}{n^2} \right)\right)
\\\\&=n+\frac{1}{2}+O\left(\frac{1}{n} \right)
\end{align}
$$ giving, as $n \to \infty$,
$$
\frac{n^{n+1}.e}{(n+1)^n}-n=\frac{1}{2}+O\left(\frac{1}{n} \right) \to L=\frac12.
$$
A: We want to compute:
$$L=\lim_{n\to +\infty} n\cdot\left(e\cdot\left(1-\frac{1}{n+1}\right)^n-1\right)\tag{1}$$
where:
$$ \log\left(1-\frac{1}{n+1}\right) = -\frac{1}{n}+\frac{1}{2n^2}+O\left(\frac{1}{n^3}\right) \tag{2}$$
gives, by multiplying by $n$ and exponentiating both terms,
$$ \left(1-\frac{1}{n+1}\right)^n = \frac{1}{e}\left(1+\frac{1}{2n}+O\left(\frac{1}{n^2}\right)\right)\tag{3}$$
now we may simply multiply both terms by $e$, subtract $1$ and multiply by $n$ to get that the value of our limit is $L=\color{red}{\large\frac{1}{2}}$.
A: We have
\begin{align}
L &= \lim_{n \to \infty}\left(\frac{n^{n + 1}e}{(n + 1)^{n}} - n\right)\notag\\
&= \lim_{n \to \infty}n\left(\exp\left(1 - n\log\left(1 + \frac{1}{n}\right)\right) - 1\right)\notag\\
&= \lim_{n \to \infty}n\left(1 - n\log\left(1 + \dfrac{1}{n}\right)\right)\dfrac{\exp\left(1 - n\log\left(1 + \dfrac{1}{n}\right)\right) - 1}{1 - n\log\left(1 + \dfrac{1}{n}\right)}\notag\\
&= \lim_{n \to \infty}n\left(1 - n\log\left(1 + \dfrac{1}{n}\right)\right)\notag\\
&= \lim_{x \to 0^{+}}\dfrac{x - \log(1 + x)}{x^{2}}\text{ (putting }x = 1/n)\notag\\
&= \lim_{x \to 0^{+}}\dfrac{x - \left(x - \dfrac{x^{2}}{2} + o(x^{2})\right)}{x^{2}}\notag\\
&= \frac{1}{2}\notag
\end{align}
We have used Taylor series for $\log(1 + x)$ at the end of the above limit evaluation and it is also possible to use L'Hospital's Rule instead of Taylor series.
