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

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. 
