$$\begin{align*}\lim_{x \to x_0} f(x)=c \Rightarrow \\ \\ \forall \epsilon > 0~~ \exists \delta > 0 ~\ni &\left|x-x_0\right|<\delta \rightarrow \left|f(x)-c \right|<\epsilon \end{align*}$$

i.e. I can choose x such that I can make f(x) arbitrarily close to the limit point.

f isn't even defined on -1 but it doesn't matter. I can find f(x) that comes arbitrarily close to 1 and thus

$$\displaystyle{\lim_{x \to -1}}f(x)=1$$

f does have a value at 0 but that value is not the limit of the function. It should be clear that I can find f(x) arbitrarily close to 0 and thus

$$\displaystyle{\lim_{x \to 0}}f(x)=0$$

at x=1 there is no limit. If I approach 1 from below f(x)=-1, if I approach it from above f(x)=0 in order for the limit to exist these limits from either side must agree.

the point x=2 is the same as the point x=-1. The function isn't defined but the limit exists and is 1.

Continuity fails when there is either no limit or the value of the function at a point is not equal to the limit.

Continuity fails at all these points.

The removable points of discontinuity are the ones where the limit exists and thus are -1, 0, and 2