Evaluating $\lim_{n \rightarrow \infty} \left( \frac{2}{\sqrt{n^2 + 4n} - n} \right) ^{B(n+2)}$ I want to solve the following problem:

Find $B \in \mathbb{R}$ such that
$$\lim_{n \rightarrow \infty} \left( \frac{2}{\sqrt{n^2 + 4n} - n} \right) ^{B(n+2)} \in \left] \frac{1}{2}, 2 \right[ \quad.$$

My attempt is as follows:
$$ \left( \frac{2}{\sqrt{x^2 + 4x} - x} \right) ^{B(x+2)} = \frac{2^{B(x+2)}}{2^{B(x+2)\log_{2} \left( \sqrt{x² + 4x} - x \right)}} = 2^{B(x+2) \left(1- \log_{2} \left( \sqrt{x² + 4x} - x \right) \right)}.$$
Thus,
$$\lim_{n \rightarrow \infty} \left( \frac{2}{\sqrt{n^2 + 4n} - n} \right) ^{B(n+2)} = 2^{B \lim_{x \rightarrow \infty}(x+2) \left(1- \log_{2} \left( \sqrt{x² + 4x} - x \right) \right)} $$
and I'm left with
$$\lim_{x \rightarrow \infty}(x+2) \left(1- \log_{2} \left( \sqrt{x² + 4x} - x \right) \right) = \lim_{x \rightarrow \infty} \frac{1- \log_{2} \left( \sqrt{x² + 4x} - x \right)}{\frac{1}{x+2}}$$
that I can compute using L'Hôpital's rule and really boring computations, arriving at
$$\tag{*} \lim_{x \rightarrow \infty} \frac{1- \log_{2} \left( \sqrt{x² + 4x} - x \right)}{\frac{1}{x+2}} = \frac{1}{\log(2)}.$$
From here I'd compute a valid interval for $B$.
Question:
Computing (*) is a horrible hassle. Is there an easier way to solve this problem? 

$$\frac{2}{\sqrt{n^2+4n}-n}=1+\frac{\sqrt{n^2+4n}-n}{2n}$$
$$=1+\frac{a_n}{n}$$  where $a_n\to 1$
It follows that 
$$(1+\frac{a_n}{n})^{n+2}\to e.$$ Does this help ?
