How to evaluate $\lim _{n\to \infty }\left(\frac{\sqrt{4n^3+3n}-2n\sqrt{n-2}}{\sqrt{2n+4}}\right)$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it. Is it possible to compute this limit with the McLaurin expansion? Can you explain the method and the steps used? Thanks. (I prefer to avoid to use l'Hospital's rule.)
$$\lim _{n\to \infty }\left(\frac{\sqrt{4n^3+3n}-2n\sqrt{n-2}}{\sqrt{2n+4}}\right)$$
 A: $$\lim _{n\to \infty }\left(\frac{\sqrt{4n^3+3n}-2n\sqrt{n-2}}{\sqrt{2n+4}}\right)=$$
$$=\lim _{n\to \infty }\left(\frac{(\sqrt{4n^3+3n}-2n\sqrt{n-2})\cdot(\sqrt{4n^3+3n}+2n\sqrt{n-2})}{\sqrt{2n+4}(\sqrt{4n^3+3n}+2n\sqrt{n-2})}\right)=$$
$$=\lim _{n\to \infty }\left(\frac{4n^3+3n-4n^3+8n^2}{\sqrt{2n+4}(\sqrt{4n^3+3n}+2n\sqrt{n-2})}\right)=$$
$$=\lim _{n\to \infty }\left(\frac{3n+8n^2}{\sqrt{2n+4}(\sqrt{4n^3+3n}+2n\sqrt{n-2})}\right)=\sqrt2$$
A: To provide an alternative (may be a little expeditious but also need a little more background) to rationalization, we may use Taylor's expansion for $(1 + x)^\alpha, \alpha > 0$ at $0$ (which is also known as McLaurin expansion):
$$(1 + x)^\alpha = 1 + \alpha x + o(x). \tag{1}$$
In view of $(1)$, the expression of interest can be written as
\begin{align}
& \frac{\sqrt{4n^3+3n}-2n\sqrt{n-2}}{\sqrt{2n+4}} \\
= & \frac{2n^{3/2}\sqrt{1 + \frac{3}{4n^2}} - 2n^{3/2}\sqrt{1 - \frac{2}{n}}}{\sqrt{2n}\sqrt{1 + \frac{2}{n}}} \\
= & \sqrt{2}n\frac{\left(1 + \frac{1}{2}\frac{3}{4n^3} + o(1/n^2)\right) - \left(1 - \frac{1}{2}\frac{2}{n} + o(1/n)\right)}{1 + \frac{1}{2}\frac{2}{n} + o(1/n)} \\
= & \frac{\sqrt{2} + o(1)}{1 + o(1)} \\
\to & \sqrt{2}
\end{align}
as $n \to \infty$.
