Other ways to find a limit where the denominator produces $0$, besides factoring and cancellation? So I came across a seemingly innocent looking integral:
$$\int_{-2}^1 \frac{1}{x^2}dx$$
Now of course, when taking the antiderivative then plugging in the values, we can see that we get a nonzero/$0$ which causes problems (because the function is discontinuous). Now to solve this we can split up the integral into two limits, both tending to $0$ from either side:
$$\left(\lim_{b\to0^-} \int_{-2}^b \frac{1}{x^2}dx\right) + \left(\lim_{a\to0^+} \int_{a}^1 \frac{1}{x^2}dx\right)$$
This would, in theory, solve the problem except for one issue, the same issue: nonzero/$0$ when plugging in the numbers to the integral from the limit.
So naturally, I thought to try and factor it somehow but within a few seconds, it became obvious that it wasn't possible. So the integral $\int_{-2}^1 \frac{1}{x^2}dx$ diverges
Now I'm wondering, is there any way to solve a limit where the denominator would produce $0$, besides factoring and cancellation?
tl;dr: A limit produces $0$ in the denominator and it is unfactorable (in the sense that you cannot cancel out the bottom term). Is there any way to solve it, using another method besides factoring and cancellation?
 A: When you are taking a limit where the denominator approaches zero and the numerator does not, you get divergence, $\infty,-\infty$ or undefined.
On the other hand if both top and bottom approach zero and you can not factor and cancel, you may take derivatives and apply the L'Hospital rule.
If the limit is too complicated then a graphing instrument may help to give you some ideas. 
A: I don't see any problem that can arise from division by zero. There is not a single time where that can actually happen in this problem ($x$ never gets to zero, it only approaches it. Therefore, a division by zero situation is just not possible):
$$
\lim_{b\to0^-} \int_{-2}^b \frac{1}{x^2}dx + \lim_{a\to0^+} \int_{a}^1 \frac{1}{x^2}dx=\\
\lim_{b\to0^-}\left[-\frac{1}{x}\right]\Big|_{-2}^{b} + \lim_{a\to0^+} \left[-\frac{1}{x}\right]\Big|_{a}^{1}=\\
\lim_{b\to0^-}\left(-\frac{1}{b}-\frac{1}{2}\right) + \lim_{a\to0^+} \left(-1+\frac{1}{a}\right)=\\
$$
Since $\lim\limits_{b\to{0^-}}\left(-\frac{1}{b}\right)=\infty$ and $\lim\limits_{a\to{0^+}}\left(\frac{1}{a}\right)=\infty$, we have:
$$
\infty-\frac{1}{2}-1+\infty=\infty
$$
