n tends to infinity $\displaystyle\lim_{n\to\infty} 
  {\left(
    \frac{\sqrt {n^2+n} - 1}{n}
  \right)}^{
  \left(2 \sqrt{n^2+n} - 1
  \right)}$
Sorry for the bad format of the question I don't know much latex
I tried rationalization of the power and the base but was not able to get the answer.

When we see both the base and exponent as variable then the best approach is to take logs. Thus if $L$ is the desired limit then
\begin{align}
\log L &= \log\left\{\lim_{n \to \infty}\left(\frac{\sqrt{n^{2} + n} - 1}{n}\right)^{(2\sqrt{n^{2} + n} - 1)}\right\}\notag\\
&= \lim_{n \to \infty}\log\left(\frac{\sqrt{n^{2} + n} - 1}{n}\right)^{(2\sqrt{n^{2} + n} - 1)}\text{ (via continuity of log)}\notag\\
&= \lim_{n \to \infty}(2\sqrt{n^{2} + n} - 1)\log\left(\frac{\sqrt{n^{2} + n} - 1}{n}\right)\notag\\
&= \lim_{n \to \infty}(2\sqrt{n^{2} + n} - 1)\log\left(1 + \frac{\sqrt{n^{2} + n} - 1 - n}{n}\right)\notag\\
&= \lim_{n \to \infty}(2\sqrt{n^{2} + n} - 1)\cdot\dfrac{\sqrt{n^{2} + n} - 1 - n}{n}\cdot\dfrac{\log\left(1 + \dfrac{\sqrt{n^{2} + n} - 1 - n}{n}\right)}{\dfrac{\sqrt{n^{2} + n} - 1 - n}{n}}\notag\\
&= \lim_{n \to \infty}(2\sqrt{n^{2} + n} - 1)\cdot\dfrac{\sqrt{n^{2} + n} - 1 - n}{n}\cdot 1\notag\\
&= \lim_{n \to \infty}\left(2\sqrt{1 + \frac{1}{n}} - \frac{1}{n}\right)\cdot\dfrac{(n^{2} + n) - (1 + n)^{2}}{\sqrt{n^{2} + n} + 1 + n}\notag\\
&= \lim_{n \to \infty}2\cdot\dfrac{-1 - n}{\sqrt{n^{2} + n} + 1 + n}\notag\\
&= -2\lim_{n \to \infty}\dfrac{1 + \dfrac{1}{n}}{\sqrt{1 + \dfrac{1}{n}} + 1 + \dfrac{1}{n}}\notag\\
&= -1\notag
\end{align}
Hence $L = 1/e$. We have used the fact that the expression $$\frac{\sqrt{n^{2} + n} - 1 - n}{n}$$ tends to $0$ as $n \to \infty$ and $(1/x)\log(1 + x) \to 1$ as $x \to 0$.
