# solving $\lim\limits_{x\rightarrow\infty} \frac{(x^2-1) \sqrt{x + 2}-x^2\sqrt{x+1}}{x\sqrt{x + 1}}$

To investigate the convergence of a series I have to solve the folliwing limit:

$$\begin{equation} \lim\limits_{x\rightarrow\infty} \frac{(x^2-1) \sqrt{x + 2}-x^2\sqrt{x+1}}{x\sqrt{x + 1}} \end{equation}$$

It should be $$\frac{1}{2}$$ but i can't quite seem to get to that solution. I've tried to factor the square root out of the quotient which was:

$$\begin{equation} \lim\limits_{x\rightarrow\infty} \sqrt{\frac{((x^2-1) \sqrt{x + 2}-x^2\sqrt{x+1})^2}{x^2(x + 1)}} \end{equation}$$

Then i worked out the square in the numerator which was:

$$\begin{equation} (x^2-1)^2 (x + 2)-2x^2(x^2-1)\sqrt{x+1}\sqrt{x+2}+x^4(x+1) \end{equation}$$

I could then factor the terms and take and divide the numerator with the (x+1) from the denominator. Then i could expand the terms in the numerator wich became:

$$\begin{equation} \lim\limits_{x\rightarrow\infty}\sqrt{\frac{(x^4+x^3-3x^2-x+2)-\sqrt{16x^8+32x^7-12x^6-20x^5+8x^4}+x^4}{x^2}} \end{equation}$$

Now i can take the $$16x^8$$ out of the root and then i then looked at the terms with the highest exponent so i had:

$$\begin{equation} \lim\limits_{x\rightarrow\infty}\sqrt{\frac{x^4-4x^4+x^4}{x^2}} =\lim\limits_{x\rightarrow\infty}\sqrt{\frac{-2x^2}{x^2}} \end{equation}$$

Which could only be solved with complex numbers, so i should be wrong somewhere in my calculations, since i know that i should get $$\frac{1}{2}$$. I also checked my sollution with WolframAlpha which also gave $$\frac{1}{2}$$ so I know that the sollution i have is correct.

Would anyone know where i was wrong or how i could better solve it?

$$\lim_{x\to\infty}\frac{(x^2-1)\sqrt{x+2}-x^2\sqrt{x+1}}{x\sqrt{x+1}}=\lim_{x\to\infty}\frac{x^2\big[\sqrt{x+2}-\sqrt{x+1}\big]}{x\sqrt{x+1}}-\frac{\sqrt{x+2}}{x\sqrt{x+1}}$$
The latter goes to $$0$$ since $$\lim_{x\to\infty}\frac{\sqrt{x+2}}{x\sqrt{x+1}}=\lim_{x\to\infty}\frac1x\cdot\sqrt{\frac{x+2}{x+1}}=\lim_{x\to\infty}\frac1x\cdot\sqrt{1+\frac1{x+1}}=0$$
You are left with $$\lim_{x\to\infty}\frac{x\big[\sqrt{x+2}-\sqrt{x+1}\big]}{\sqrt{x+1}}=\lim_{x\to\infty}\frac{x\big[\sqrt{x+2}-\sqrt{x+1}\big]}{\sqrt{x+1}}\cdot\frac{\big[\sqrt{x+2}+\sqrt{x+1}\big]}{\big[\sqrt{x+2}+\sqrt{x+1}\big]}\\=\lim_{x\to\infty}\frac x{\sqrt{x+1}\big[\sqrt{x+2}+\sqrt{x+1}\big]}=\lim_{x\to\infty}\frac 1{\sqrt{1+\frac1x}\Big[\sqrt{1+\frac2x}+\sqrt{1+\frac1x}\Big]}=1/2$$
Another way: $$\begin{eqnarray*} \frac{(x^2-1) \sqrt{x + 2}-x^2\sqrt{x+1}}{x\sqrt{x + 1}} & = & \frac{(x-1)(x+1) \sqrt{x + 2}-x^2\sqrt{x+1}}{x\sqrt{x + 1}} \\ & = & \frac{(x-1)\sqrt{(x+1)(x + 2)}-x^2}{x} \\ & \stackrel{x^2 = x(x-1)+x}{=} & \underbrace{\frac{x-1}{x}}_{\stackrel{x \to +\infty}{\longrightarrow}1}\underbrace{\left(\sqrt{(x+1)(x + 2)} - x\right)}_{= \frac{3x+2}{\sqrt{(x+1)(x + 2)} + x}\stackrel{x \to +\infty}{\longrightarrow}\frac{3}{2}} - 1 \\ & \stackrel{x \to +\infty}{\longrightarrow} & 1 \cdot \frac{3}{2} -1 = \frac{1}{2} \end{eqnarray*}$$