Find the following limit: $L=\lim_{n\to \infty}\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}$ Find the following limit:
$$L=\lim_{n\to \infty}\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}$$
I think use Taylor's expansion give $\left(2\sqrt[n]{n}-1\right)^n$ or there is a workaround, but I do not know. 
 A: Let $b_n=\sqrt[n]{n}-1=n^{1/n}-1$ and $a_n=\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}$ so that
$$a_n=\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}=\frac{\left(2\sqrt[n]{n}-2+1\right)^n}{n^2}=\frac{1}{n^2}\operatorname{e}^{n\ln(1+2b_n)}$$
For $n\to \infty, \;b_n\to 0$ because $b_n=\operatorname{e}^{\frac{\ln n}{n}}-1$ and for $n\to \infty$, $\frac{\ln n}{n}\to 0$.
So for $n\to\infty$ we have $\ln(1+2b_n)\sim 2b_n$, and $b_n=n^{1/n}-1=\operatorname{e}^{\frac{\ln n}{n}}-1\sim 1+\frac{\ln n}{n}-1=\frac{\ln n}{n}$ so that
$$
n\ln(1+2b_n)\sim n2b_n\sim n2\frac{\ln n}{n}=\ln n^2.
$$
Finally for $n\to\infty$
$$
a_n=\frac{1}{n^2}\operatorname{e}^{n\ln(1+2b_n)}\sim \frac{1}{n^2}\operatorname{e}^{\ln n^2}=\frac{n^2}{n^2}=1
$$
that is
$$
\lim_{n\to\infty}\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}=1
$$
A: $$
\begin{align}
L&=\lim_{n\to \infty}\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}=\lim_{n\to \infty}\left(\frac{2\sqrt[n]{n}-1}{n^{2/n}}\right)^n \\
&= \operatorname{e}^{\lim_{n\to \infty}\left(\frac{2\cdot\sqrt[n]{n}-1}{n^{2/n}}-1\right)n}=\operatorname{e}^{-\lim_{n\to \infty}\left(\frac{n^{1/n}-1}{n^{1/n}}\right)^2\cdot n}\\
&= \operatorname{e}^{-\lim_{n\to \infty}\left(\frac{\operatorname{e}^{\frac{\ln n}{n}}-1}{{\frac{\ln n}{n}}}\right)^2\cdot\left(\frac{\ln n}{n}\right)^2\cdot \frac{1}{n^{2/n}}\cdot n}\\
&=\operatorname{e}^{-1\cdot 1\cdot \lim_{n\to \infty}\frac{\ln^2n}{n}}=\operatorname{e}^{0}= 1
\end{align}$$ 
