<s>[INST] Checking my proof of finding a limit I'm trying to find all the $a\in \mathbb R$ values which we get the following limit:
$$L=\lim_{x\to a} \frac{x^2+ax-2a}{(x^2-1)(x+a)}$$
to be finite. I don't see any problem just inserting $a$ into the limit so we get $\frac{1}{a+1}$ so for $a\neq -1$ we get a finite limit. Is it that easy?
 [/INST] In this case, as long as we don't have $\frac{0}{0}$ or $\frac{\infty}{\infty}$, yes it is that easy and inserting $a$ is enough. And the answer becomes $a \in \mathbb{R}-\{-1\}$.
EDIT: By the warning from fleablood, we should also consider the case where we can have $\frac{0}{0}$. Here, note that in order for nominator to be $0$, we must have $2a^2-2a = 0 \implies a = 0 \lor a = 1$.
Case 1: $a = 0$. In this case, we have 
$$\lim_{x \to 0}\frac{x^2}{(x^2-1)x} = \lim_{x \to 0}\frac{x}{(x^2-1)} = 0 \ne \frac{1}{a+1} = 1$$
Case 2: $a = 1$. In this case, we have
$$\lim_{x \to 1}\frac{x^2+x-2}{(x^2-1)(x+1)} = \lim_{x \to 1}\frac{(x+2)(x-1)}{(x^2-1)(x+1)} =  \lim_{x \to 1}\frac{(x+2)}{(x+1)^2} = \frac{3}{4} \ne \frac{1}{a+1} = \frac{1}{2}$$
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