Please help me solve this question:
lim (x^3-3x-2)/
x->2 (x^2-5x+6)
$\displaystyle \lim_{x \to 2} \frac{x^3 - 3x - 2}{x^2 - 5x + 6}$
Note that $\displaystyle x - 2$ is a factor of $\displaystyle x^3 - 3x - 2$, evidenced by the fact that $\displaystyle 2^3 - 3 \cdot 2 - 2 = 0$.
Thus, by long or synthetic division we get that
$\displaystyle x^3 - 3x - 2 = (x - 2)(x + 1)^2$
The denominator factors.
Thus
$\displaystyle \lim_{x \to 2} \frac{x^3 - 3x - 2}{x^2 - 5x + 6} = \lim_{x \to 2} \frac{(x - 2)(x + 1)^2}{(x - 2)(x - 3)}$
$\displaystyle = \lim_{x \to 2} \frac{(x + 1)^2}{x - 3}$
$\displaystyle = \frac{9}{-1} = -9$
-Dan