# limited thoughts on limits :S

• Oct 7th 2007, 07:57 AM
redpanda11
limited thoughts on limits :S

lim (x^3-3x-2)/
x->2 (x^2-5x+6)
• Oct 7th 2007, 08:02 AM
Jhevon
Quote:

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)
• Oct 7th 2007, 08:06 AM
topsquark
Quote:

Originally Posted by redpanda11

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