# Limits

• Dec 7th 2006, 10:53 PM
Limits
Hi how can i factor this limit?

Find the limit
$\displaystyle \lim_{x \to -1} \frac{2x^2 - x - 3}{x^3+2x^2+6x+5}$
• Dec 7th 2006, 11:35 PM
parallel
I get an error when using latex so

to factor this limit:

2x^2 - x - 3 = (2x-3)(x+1)

x^3 + 2x^2 + 6x + 5 = (x+1)(x^2+x+5)

the term (x+1) cancels out,and you can compute the limit.
• Dec 7th 2006, 11:42 PM
THanks you very much parallel

how did you factor the denominator?
• Dec 8th 2006, 04:25 AM
topsquark
Quote:

THanks you very much parallel

how did you factor the denominator?

Do you know the Rational Roots Theorem?

Given a polynomial:
$\displaystyle px^n + ax^{n-1} + ... + bx + q = 0$
If a rational root of this polynomial exists it will be of the form:
$\displaystyle x = \frac{\text{factor of q}}{\text{factor of p}}$

In this case we have:
$\displaystyle x^3 + 2x^2 + 6x + 5$

So if there is a factor of this over the rational numbers it will be of the form:
$\displaystyle (\text{factor of 1}) x - (\text{factor of 5})$

So the possible rational factors of $\displaystyle x^3 + 2x^2 + 6x + 5$ are:
$\displaystyle (x \pm 1)$
$\displaystyle (x \pm 5)$

You can try each of these and find that (x + 1) is the only linear rational factor.

-Dan