# 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.)

$$\lim _{n\to \infty }\left(\frac{\sqrt{4n^3+3n}-2n\sqrt{n-2}}{\sqrt{2n+4}}\right)$$

• if you see ever square roots, then just rationalize! – user5954246 May 11 '16 at 14:43

## 2 Answers

$$\lim _{n\to \infty }\left(\frac{\sqrt{4n^3+3n}-2n\sqrt{n-2}}{\sqrt{2n+4}}\right)=$$ $$=\lim _{n\to \infty }\left(\frac{(\sqrt{4n^3+3n}-2n\sqrt{n-2})\cdot(\sqrt{4n^3+3n}+2n\sqrt{n-2})}{\sqrt{2n+4}(\sqrt{4n^3+3n}+2n\sqrt{n-2})}\right)=$$ $$=\lim _{n\to \infty }\left(\frac{4n^3+3n-4n^3+8n^2}{\sqrt{2n+4}(\sqrt{4n^3+3n}+2n\sqrt{n-2})}\right)=$$ $$=\lim _{n\to \infty }\left(\frac{3n+8n^2}{\sqrt{2n+4}(\sqrt{4n^3+3n}+2n\sqrt{n-2})}\right)=\sqrt2$$

• I thank you but the exact result is $\sqrt2$ – Amarildo May 11 '16 at 14:43

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 $n \to \infty$.

• Ah, you actually mentioned using McLaurin expansion, then I think this is what exactly you need. – Zhanxiong May 11 '16 at 15:40