The original poster asked in a comment: "What if we can't factorize? Should we assume then that the limit DNE? or is 0?"
You had $\lim_{x \to 2} \frac{3x^2-x-10}{x^2-4}$ and as $x\to2$, the numerator and denominator both approach $0$. If the numerator and denominator are polynomials, then:
- If the numerator approaches a non-zero number and the denominator approaches $0$, then the limit does not exist (in some cases it's $\infty$ or $-\infty$, in which case the result is often phrased as "the limit does not exist").
- If the denominator is a non-zero number then just plug in the number that $x$ is approaching (in this case $2$) and that's the limit.
- (The really important case) If they both approach $0$, then use the fact from algebra described below.
Algebra tells us that if you plug a number into a polynomial and get $0$, that tells you something about how to factor it. If you plug $2$ into $3x^2-x-10$ and get $0$, that means $x-2$ is one of the factors. Similarly, you plug $2$ into $x^2-4$ and get $0$, and that tells you $x-2$ is one of the factors. So $ \frac{3x^2-x-10}{x^2-4} = \frac{(x-2)(\cdots\cdots)}{(x-2)(\cdots\cdots)}. $ Then you need to figure out what goes in place of $(\cdots\cdots)$ in each case. You can do that by long division, dividing $3x^2-x-10$ by $x-2$, and similarly dividing $x^2-4$ by $x-2$. That will work even in cases that would be difficult to factor if you didn't have this way of knowing that $x-2$ is one of the factors. And you don't need to factor completely; you only need to pull out any factors that are $0$ when $x=\text{(in this case) }2$.