<s>[INST] 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.
 [/INST] 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 $x \to 0$.
</s>