# Math Help - limited thoughts on limits :S

1. ## limited thoughts on limits :S

lim (x^3-3x-2)/
x->2 (x^2-5x+6)

2. Originally Posted by redpanda11

lim (x^3-3x-2)/
x->2 (x^2-5x+6)
factorize the numerator and denominator and your problem will cancel out. (note that (x - 2) is a factor of the cubic in the numerator)

3. Originally Posted by redpanda11

lim (x^3-3x-2)/
x->2 (x^2-5x+6)
$\lim_{x \to 2} \frac{x^3 - 3x - 2}{x^2 - 5x + 6}$

Note that $x - 2$ is a factor of $x^3 - 3x - 2$, evidenced by the fact that $2^3 - 3 \cdot 2 - 2 = 0$.

Thus, by long or synthetic division we get that
$x^3 - 3x - 2 = (x - 2)(x + 1)^2$

The denominator factors.

Thus
$\lim_{x \to 2} \frac{x^3 - 3x - 2}{x^2 - 5x + 6} = \lim_{x \to 2} \frac{(x - 2)(x + 1)^2}{(x - 2)(x - 3)}$

$= \lim_{x \to 2} \frac{(x + 1)^2}{x - 3}$

$= \frac{9}{-1} = -9$

-Dan