Why this limit can't be computed like this? This is the limit:
$$\lim_{x \to 2} \frac{3x^2-x-10}{x^2-4}$$
The solution manual says it is: $\frac{11}{4}$
I've tried to solve it like a polynomial like this:
$$ \frac{(\frac{3x^2}{3x^2}-\frac{x}{3x^2}-\frac{10}{3x^2})*3x^2}{(\frac{x^2}{x^2}-\frac{4}{x^2})*x^2}=$$
$$= \frac{(1-0-0)*3x^2}{(1+0)*x^2}=3$$
Can you please tell me where am I doing wrong? Or why these kind of operation doesn't work here? Thank you
 A: For this problem, you'll want to factor the numerator and the denominator, (which both tend to $0$ as $ x \to 2$), cancel a factor of $x-2$, and then try direct substitution again. 
$\frac{3x^2 -x - 10}{x^2-4} = \frac{(3x +5)(x - 2)}{(x + 2)(x - 2)} = \frac{3x + 5}{x + 2}$ (when $x \neq 2$)
Now letting $x \to 2$, we get $\frac{3 \cdot 2 + 5}{2 + 4} = \frac{11}{4}$, as desired. 
A: 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 $x=\text{(in this case) }2$.
