$$\lim_{x\to\infty}\frac{(x^2-1)\sqrt{x+2}-x^2\sqrt{x+1}}{x\sqrt{x+1}}=\lim_{x\to\infty}\frac{x^2\big[\sqrt{x+2}-\sqrt{x+1}\big]}{x\sqrt{x+1}}-\frac{\sqrt{x+2}}{x\sqrt{x+1}}$$
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 $$\lim_{x\to\infty}\frac{x\big[\sqrt{x+2}-\sqrt{x+1}\big]}{\sqrt{x+1}}=\lim_{x\to\infty}\frac{x\big[\sqrt{x+2}-\sqrt{x+1}\big]}{\sqrt{x+1}}\cdot\frac{\big[\sqrt{x+2}+\sqrt{x+1}\big]}{\big[\sqrt{x+2}+\sqrt{x+1}\big]}\\=\lim_{x\to\infty}\frac x{\sqrt{x+1}\big[\sqrt{x+2}+\sqrt{x+1}\big]}=\lim_{x\to\infty}\frac 1{\sqrt{1+\frac1x}\Big[\sqrt{1+\frac2x}+\sqrt{1+\frac1x}\Big]}=1/2$$
It is meaningless state that $\lim\limits_{x\to \infty} \sqrt{x^2+x}-x \ne x - x$ and $\lim\limits_{x\to \infty} \sqrt{x^2+x}+x = x + x$ we should state that
$$\lim\limits_{x\to \infty} \sqrt{x^2+x}-x = \lim\limits_{x\to \infty} \frac1{2}+o(1/x)=\frac12$$
and
$$\lim\limits_{x\to \infty} \sqrt{x^2+x}+x = \lim\limits_{x\to \infty} 2x+o(1)=\infty$$
the explanation in both case is in binomial first order approximation that is
$$\sqrt{x^2+x}=x\left(1+\frac1x\right)^\frac12= x\left(1+\frac1{2x}+o\left(\frac1x\right)\right)=x+ \frac1{2}+o\left(1\right) $$
which means that for $x$ large we have
$$\sqrt{x^2+x}\sim x+ \frac1{2}$$
and therefore
$$\sqrt{x^2+x}-x \sim \frac 12$$
$$\sqrt{x^2+x}+x \sim 2x+\frac 12$$
Edit
Note that for $\frac{x}{ \sqrt{x^2+x} + x}$ it is not correct to state that the denominator is $2x$ the complete steps are
$$\frac{x}{ \sqrt{x^2+x} + x}=\frac x x \frac{1}{ \sqrt{1+1/x} + 1} \to \frac12$$
For $\sqrt{x^2+x} - x$ indeed is not correct take$ \sqrt{x^2+x}=x$ what is true is that $\sqrt{x^2+x}\sim x+\frac12$.
If we use that approximation, it works with both limits. In this particular case first order approximation works and we can use it to evaluate both limits.
Best Answer
What you did wrong:
for $x<0, \sqrt{x^2}=-x$