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.
 A: We have
$$\frac{\sqrt{n^2+n}-1}{n}=\sqrt{1+1/n}-1/n=1+\frac1{2n}+o\left(\frac1{n}\right)-\frac1n=1-\frac1{2n}+o\left(\frac1{n}\right)$$
then
$$\left(\frac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}=\left[\left(1-\frac1{2n}+o\left(\frac1{n}\right)\right)^{\frac1{-\frac1{2n}+o\left(\frac1{n}\right) }}\right]^{(2\sqrt{n^2+n}-1)\left(-\frac1{2n}+o\left(\frac1{n}\right)\right)}\to \frac1e$$
indeed
$$(2\sqrt{n^2+n}-1)\left(-\frac1{2n} +o\left(\frac1{n}\right)\right)=-\frac{\sqrt{n^2+n}}{n}+o(1)\to -1$$
A: By standard limits we have that
$$\left(\frac{\sqrt{n^2+n}-1}{n}\right) ^ {2\sqrt{n^2+n}-1}=e^{({2\sqrt{n^2+n}-1})\log \left(\frac{\sqrt{n^2+n}-1}{n}\right)} \to \frac1e$$
since
$${{(2\sqrt{n^2+n}-1)}\log \left(\frac{\sqrt{n^2+n}-1}{n}\right)} \to -1$$
indeed
$$(2\sqrt{n^2+n}-1)\log \left(\frac{ \sqrt{n^2+n}-1}{n}\right)=\\
=(2\sqrt{n^2+n}-1)\left[\log\left(\frac{ \sqrt{n^2+n}}{n}\right)+\log \left(1-\frac1{ \sqrt{n^2+n}}\right)\right]=\\
=(2\sqrt{n^2+n}-1)\left[\frac12\log\left(1+\frac1n\right)+\log \left(1-\frac1{ \sqrt{n^2+n}}\right)\right]=\\
=\left(2 -\frac1{\sqrt{n^2+n}}\right)\left[\frac12\log\left(1+\frac1n\right)^{\sqrt{n^2+n}}+\log \left(1-\frac1{ \sqrt{n^2+n}}\right)^{\sqrt{n^2+n}}\right]=\\
=\left(2 -\frac1{\sqrt{n^2+n}}\right)\left[\frac12\log\left[\left(1+\frac1n\right)^{n}\right]^{\sqrt{1+\frac1n}}+\log \left(1-\frac1{ \sqrt{n^2+n}}\right)^{\sqrt{n^2+n}}\right]\to2\left(\frac12\cdot1-1\right)=-1
$$
A: $$
\left(\sqrt{1+\frac{1}{n}}-\frac{1}{n}\right)^{2n\left(\sqrt{1+\frac{1}{n}}-\frac{1}{2n}\right)} = \left(\sqrt{1+\frac{1}{n}}-\frac{1}{n}\right)^{2n+O(\frac{1}{n})}
$$
hence
$$
\lim_{n\to\infty}\left(\sqrt{1+\frac{1}{n}}-\frac{1}{n}\right)^{2n+O(\frac{1}{n})} = e^{-1}
$$
because
$$
\lim_{y\to 0}\left(\sqrt{1-y}-y\right)^{\frac{2}{y}} = \lim_{y\to 0}\left(1-\frac{y}{2}+O(y^2)\right)^{\frac{2}{y}} = e^{-1}
$$
A: 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 $0^0$.)
