<s>[INST] Finding $\lim_{n\to\infty} \left(\frac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}$ 
Find the following limit without using the L'Hopital rule:
  $$\lim_{n\to\infty} \left(\dfrac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}$$
Answer: $e^{-1}$

My attempt: Since the limit is of the form $1^{\infty}$, I decided to use the standard formula:

$$\lim_{x\to a} f^g = e^{\lim_\limits{x\to a}(f-1)g}$$

(See link)
Let $l=(f-1)g$.
We have, 
$$l=\left(\dfrac{\sqrt{n^2+n}-(1+n)}{n}\right)(2\sqrt{n^2+n}-1)$$
This on solving boils down to
$$l=2n+3-\sqrt{1+\frac 1n}(2n+3)+\frac 1n$$
Now if I tend $n$ towards infinity, then $l\to 0$ and the limit i.e. $e^l$, is equal to $1$, which contradicts the given answer. 
Please help. Thanks!
Edit: 
A proof for the "standard formula" I have used.

Edit 2: just noticed a typo in the power which I have now fixed.
 [/INST] $ \lim_{n \to \infty} l = -1 $, not $ 0 $. Indeed,
\begin{align}
2n + 3 - \sqrt{1 + \frac{1}{n}}(2n + 3) + \frac{1}{n}
&= \left(1 - \sqrt{1 + \frac{1}{n}}\right)(2n + 3) + \frac{1}{n} \\
&= \frac{(1 - \sqrt{1 + 1/n})(1 + \sqrt{1 + 1/n})}{1 + \sqrt{1 + 1/n}} \cdot (2n + 3) + \frac{1}{n} \\
&= \frac{-1/n}{1 + \sqrt{1 + 1/n}} \cdot (2n + 3) + \frac{1}{n} \\
&= - \frac{2 + 3/n}{1 + \sqrt{1 + 1/n}} + \frac{1}{n} \\
&\to -1
\end{align}
as $ n \to \infty $.
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