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.

Let $y = \displaystyle\lim_{n\to\infty} 
  {\left(
    \frac{\sqrt {n^2+n} - 1}{n}
  \right)}^{2 \sqrt{n^2+n} - 1}$
and let $g(n) = \sqrt{n^2 + n}$
Then $g'(n) = \dfrac{2n+1}{2g(n)}$ and
\begin{align}
    \ln(y) 
    &= (2\sqrt{n^2+n} - 1) \ln{ \left( \frac{\sqrt{n^2+n} - 1}{n} \right)} \\
    &= (2g(n) - 1) \ln{ \left( \frac{g(n) - 1}{n} \right)} \\
    &= \dfrac
       {ln(g(n) - 1) -\ln n}
       {\left( \dfrac{1}{2 g(n) + 1} \right)}
\end{align}
Using L'Hospital,
\begin{align}
    \lim_{n\to\infty}\dfrac
       {ln(g(n) - 1) -\ln n}
       {\left( \dfrac{1}{2 g(n) + 1} \right)}
    &=-\lim_{n\to\infty}\dfrac
       {\left( \dfrac{g'(n)}{g(n)-1} \right) - \dfrac 1n}
       {\left( \dfrac{2g'(n)}{(2g(n)+1)^2} \right)} \\
    &=-\lim_{n\to\infty}\dfrac
       {\left( \dfrac{1}{g(n)-1} \right) - \dfrac{2g(n)}{n(2n+1)}}
       {\left( \dfrac{2}{4n^2 + 4n + 4g(n) + 1} \right)} \\
    &=-\lim_{n\to\infty}\dfrac
       {\left( \dfrac{2n^2 + n - 2n^2 - 2n + 2g(n)}{(2n^2 + n)(g(n)-1)} \right)}
       {\left( \dfrac{2}{4n^2 + 4n + 4g(n) + 1} \right)} \\
    &=-\lim_{n\to\infty}
      \dfrac
          {(-n + 2g(n))(4n^2 + 4n + 4g(n) + 1)}
          {2(2n^2 + n)(g(n)-1)} \\
    &=-\lim_{n\to\infty}
      \dfrac
          {2\sqrt{n^2 + n}-n}
          {\sqrt{n^2 + n}-1}
      \cdot \lim_{n\to\infty} \dfrac
          {(4n^2 + 4n + 4g(n) + 1)}
          {4n^2 + 4n} \\
    &= -1
\end{align}
So $\ln y = -1$ and $y = \dfrac 1e$
