Finding $\lim\limits_{n→∞}\left(\frac{\sqrt{n^2+n}-1}n\right)^{2\sqrt{n^2+n}-1}$ I have the limit
$$\lim_{n\to\infty}\left(\frac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}.$$
I simplified the limit by using $(f(x)-1)\cdot g(x)$, where $f(x)= \sqrt{n^2+n}-1$  and $g(x) = 2\sqrt{n^2+n}- 1$ since it is in the form of $1^\infty$. I ended up getting $$\frac{(3+2n)(n-(\sqrt{n^2-n})+1}{n},$$ which tends to $\infty$.
Please help how to proceed further or a different way to solve it.

First of all, note that the $-1$ in the exponent is irrelevant, i.e.,
$$\lim_{n\to\infty}\left(\sqrt{n^2+n}-1\over n \right)^{2\sqrt{n^2+n}-1}
=\lim_{n\to\infty}\left(\left(\sqrt{n^2+n}-1\over n \right)^{2\sqrt{n^2+n}}\over\sqrt{1+{1\over n}}-{1\over n}\right)
=\lim_{n\to\infty}\left(\sqrt{n^2+n}-1\over n \right)^{2\sqrt{n^2+n}}$$
Now, abbreviating $\sqrt{n^2+n}$ to $s$ (for "$s$quare root") when convenient, and noting that $s\to\infty$ as $n\to\infty$, we have
$$\begin{align}
\left(\sqrt{n^2+n}-1\over n \right)^{2\sqrt{n^2+n}}
&=\left(\sqrt{n^2+n}-1\over\sqrt{n^2+n} \right)^{2\sqrt{n^2+n}}\left(\sqrt{n^2+n}\over n \right)^{2\sqrt{n^2+n}}\\
&=\left(1-{1\over s} \right)^{2s}\left(n^2+n\over n^2 \right)^{\sqrt{n^2+n}}\\
&=\left(\left(1-{1\over s} \right)^s\right)^2\left(\left(1+{1\over n} \right)^n\right)^{\sqrt{1+1/n}}\\
&\to (e^{-1})^2\cdot (e^1)^1=e^{-1}
\end{align}$$
(Note, in the limiting step, we use both the product theorem for limits and the theorem that $f(n)^{g(n)}\to L^\ell$ if $f(n)\to L$ and $g(n)\to\ell$ and $L^\ell$ is not of the indeterminate form $0^0$.)
