Four limits with absolute value and trigonometry

a) For $x\neq0$, $\dfrac{x^2+2|x|}{x}=x+2\dfrac{|x|}{x}$.
- On $x>0$: $|x|/x=1$, so the expression equals $x+2\to2$.
- On $x<0$: $|x|/x=-1$, so the expression equals $x-2\to-2$.

Left and right limits differ, so the limit does not exist.

b) For $x<0$, $|x|=-x$, so
$$\frac{x^2-2x}{x}=x-2\longrightarrow-\infty.$$
The limit is $-\infty$.

c) Factor: $x^2-4=(x-2)(x+2)$, $x^2-3x+2=(x-2)(x-1)$. For $x\neq2$:
$$\frac{x+2}{x-1}\longrightarrow\frac{2+2}{2-1}=4.$$
The limit is $4$.

d) Factor $\sin^2 x=(1-\cos x)(1+\cos x)$. For $\cos x\neq-1$:
$$\frac{\sin^2 x}{1+\cos x}=1-\cos x\longrightarrow 1-(-1)=2.$$
The limit is $2$.