Evaluate the algebraic limit $\lim_{n\to\infty} \frac{2n}{\sqrt{n+2}(\sqrt{n+1} + \sqrt{n-1})} $ 
$$\lim_{n\to\infty} \frac{2n}{\sqrt{n+2}(\sqrt{n+1} + \sqrt{n-1})} $$

I know that its value is somewhere in $(0, \infty)$ but I have no idea how to find the exact value.
Update: I've had a mistake in formulation of the limit in the original question so I updated the limit.
 A: Note: This answer refers to the original question that OP asked before edited the question to the new one.
Dividing by $n$, we get,
$$\lim_{n\to\infty} \frac{2n}{\sqrt{n+1} + \sqrt{n-1}} =\lim_{n\to \infty}\frac{2}{\sqrt{\frac{1}{n^2}+\frac{1}{n}}+\sqrt{\frac{1}{n}-\frac{1}{n^2}}}$$
So the limit tends to $\infty$ when $n$ tends to $\infty$, since the denominator tends to $0$.
A: \begin{align*}
\lim_{n\to\infty} \dfrac{2n}{\sqrt{n+1}+\sqrt{n-1}}&=\lim_{n\to\infty} \dfrac{2n \cdot \frac{1}{\sqrt{n}}}{\left(\sqrt{n+1}+\sqrt{n-1}\right)\cdot \frac{1}{\sqrt{n}}}\\
&=\lim_{n\to\infty} \dfrac{2\sqrt{n}}{\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}}\\
&=+\infty
\end{align*}
because the numerator tends to infinity while the denominator tends to 2.
[A good technique when solving a limit of "rational functions" is to divide both the denominator and the numerator by the greatest power of the denominator, which in this case is intuitively $\sqrt{n}$]
A: $$\frac{2n}{\sqrt{n+1} + \sqrt{n-1}}$$
$$=\frac{1}{\sqrt{n}}\frac{1}{\sqrt{1+\frac{1}{n}} + \sqrt{1-\frac{1}{n}}}$$
$$=2\sqrt{n} \cdot \frac{1}{1+\frac{1}{2n}-\frac{1}{8n^2} + 1-\frac{1}{2n}-\frac{1}{8n^2} + o\left(\frac{1}{n^2}\right)}$$
$$=2\sqrt{n} \cdot \frac{1}{2-\frac{1}{4n^2}+o\left(\frac{1}{n^2}\right)}$$
$$=\sqrt{n} \cdot \left(1+\frac{1}{8n^2}+o\left(\frac{1}{n^2}\right)\right)\underset{n\to +\infty}{\longrightarrow}\infty$$
A: Note: This is for the new question. 
Again dividing by $n$, we get,
$$\lim_{n\to\infty} \frac{2n}{\sqrt{n+2}(\sqrt{n+1} + \sqrt{n-1})} =\lim_{n\to \infty}\frac{2}{\sqrt{1+\frac{2}{n}}\left(\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}\right)}=1$$
