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.
 A: Let $\sqrt{n^2+n}=t$, then $n^2+n=t^2$ from which through completing the square we find $n=-0.5+\sqrt{t^2+0.25}$. I picked the positive root because of the limit of $n$ going to positive infinity. Inserting this substitution into the limit we arrive at the expression
$$\left(\frac{t-1}{-0.5+\sqrt{t^2-0.25}}\right)^{(2t-1)}$$
where $t$ goes to infinity. Next substitution: $t=0.5\tan v$. Observe that now $v$ goes to $\pi/2$ (from the left side, but that makes no difference here for the outcome of the limit). The new limit expression becomes 
$$\left(\frac{\tan v-2}{-1+\sec v}\right)^{(\tan v-1)}$$
At this point we can call this expression $y$ and it is time to take the $\ln$ on both sides, as you also indicated in your post. Using the property of the $\ln$ by taking the exponent up front, we arrive at
$$(\tan v-1)\ln\left(\frac{\tan v-2}{\sec v-1}\right)$$
Now using sines and cosines instead and multiply through by $\cos v$ to take out denominators in the $\ln$; verify you get:
$$(\sin v-\cos v)\frac{\ln(\sin v-2\cos v)-\ln(1-\cos v)}{\cos v}$$
You can now use Hospital's Rule on the second term because it is a zero over zero situation. For that reason I split the $\ln$ so that taking derivatives becomes easier. The derivative of that cosine term in the denominator is $-\sin v$ which can be brought as a denominator in front. so we get
$$\frac{\sin v-\cos v}{-\sin v}\left(\frac{\cos v+2\sin v}{\sin v-2\cos v}-\frac{\sin v}{1-\cos v}\right)$$
Plugging in $v=\pi/2$ results in
$$\frac{1}{-1}\left(\frac{0+2}{1-0}-\frac{1}{1-0}\right) = -1.$$
Since this is $\ln y$, it follows that the answer is $e^{-1}$ which is approximately $0.3678\ldots$ which I verified with DESMOS by graphing your original limit.
A: 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 $y = \dfrac 1e$
A: Let $f$ and $g$ be functions such that $\lim_{x\rightarrow a}f=1$ and $\lim_{x\rightarrow a}g=\infty$.
Let $$L=\lim_{x\rightarrow a}f^g$$
$$\log L=\log(\lim_{x\rightarrow a}f^g)$$
$$\log L=\lim_{x\rightarrow a}\log(f^g)$$
$$\log L=\lim_{x\rightarrow a}g\log(f)$$
$$L=e^{\lim_{x\rightarrow a}g\log(f)}$$
$$L=e^{\lim_{x\rightarrow a}g\frac{\log(f)}{f-1}(f-1)}$$
$$L=e^{\lim_{x\rightarrow a}g\lim_{x\rightarrow a}\frac{\log(f)}{f-1}\lim_{x\rightarrow a}(f-1)}$$
$$\lim_{x\rightarrow a}\frac{\log(f)}{f-1}=1$$
Thus,
$$L=e^{\lim_{x\rightarrow a}g(f-1)}$$
Using this in the given question,
$$\lim_{n\to\infty}{\left(\frac{\sqrt {n^2+n} - 1}{n}\right)}^{\left(2\sqrt{n^2+n} - 1\right)}=e^{\lim_{n\to\infty}\left(2\sqrt{n^2+n} - 1\right)\left(\frac{\sqrt {n^2+n} - 1}{n}-1\right)}=e^{\lim_{n\to\infty}\frac{\left(2\sqrt{n^2+n} - 1\right)}{n}\left(\sqrt {n^2+n} -1- n\right)}=e^{\lim_{n\to\infty}\frac{\left(2\sqrt{n^2+n} - 1\right)}{n}\left(\sqrt {n^2+n} -1- n\right)}=e^{\lim_{n\to\infty}\frac{\left(2\sqrt{n^2+n} - 1\right)}{n}\lim_{n\to\infty}\left(\sqrt {n^2+n} -1- n\right)}=e^{\lim_{n\to\infty}\frac{\left(2n\sqrt{1+\frac1n} - 1\right)}{n}\lim_{n\to\infty}\left(n\sqrt {1+\frac1n} -1- n\right)}=e^{2\lim_{n\to\infty}\left(n(\sqrt {1+\frac1n}-1) -1\right)}=e^{2\lim_{n\to\infty}\left(n(\sqrt {1+\frac1n}-1)\frac{(\sqrt {1+\frac1n}+1)}{(\sqrt {1+\frac1n}+1)} -1\right)}=e^{2\lim_{n\to\infty}\left(n\frac{\frac1n}{(\sqrt {1+\frac1n}+1)} -1\right)}=e^{2(\frac12 -1)}=e^{-1}$$
A: Let $1/x=h\implies h\to0^+,h>0$ 
$$F=\lim_{n\to\infty} 
  {\left(    \frac{\sqrt {n^2+n} - 1}{n}  \right)}^{  \left(2 \sqrt{n^2+n} - 1  \right)}$$
$$ =\left[\lim_{h\to0^+}\left(1+\sqrt{1+h}-h-1\right)^{\frac1{\sqrt{1+h}-h-1}}\right]^{\lim_{h\to0^+}\dfrac{(\sqrt{1+h}-h-1)}h\cdot\lim_{h\to0^+}(2\sqrt{1+h}-1)}$$
Clearly, the inner limit converges to $e$
For $G=\lim_{h\to0^+}\dfrac{(\sqrt{1+h}-h-1)}h,$  set $\sqrt{1+h}=u+1\implies h=u^2+2u$
$G=\lim_{u\to0}\dfrac{(u+1)-(u+1)^2}{u^2+2u}=-\lim_{u\to0}\dfrac{u+1}{u+2}=?$
$\lim_{h\to0^+}(2\sqrt{1+h}-1)=?$
A: 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 $x \to 0$.
