1. Jun 4, 2004

Thanks...

Evaluate the limit

lim x->-1 (2x^2-x-3)/(x+1)

I simplified to (2x^2-3)/(2x+1) = -1/-1 = 1

Is this correct?

2. Jun 4, 2004

Zorodius

I believe you want to factor the top, cancel out the hole in the graph, and then evaluate by substitution.

$$\lim_{x\rightarrow -1} \frac {2x^2-x-3}{x+1}$$

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

$$= \lim_{x\rightarrow -1} 2x-3$$

= -5.

3. Jun 4, 2004

Caldus

You can use L'Hospital's rule:

$$\lim_{x\rightarrow -1} \frac {2x^2-x-3}{x+1}$$

If you plug in, you get 0/0, so:

$$\lim_{x\rightarrow -1} \frac {2x^2-x-3}{x+1}$$
= $$\lim_{x\rightarrow -1} \frac {4x-1}{1}$$

Plug in and you get -5.