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

$$\lim_{x\to\infty}\frac{(x^2-1)\sqrt{x+2}-x^2\sqrt{x+1}}{x\sqrt{x+1}}=\lim_{x\to\infty}\frac{x^2\big[\sqrt{x+2}-\sqrt{x+1}\big]}{x\sqrt{x+1}}-\frac{\sqrt{x+2}}{x\sqrt{x+1}}$$

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*}