How about this limit?

ladyrae

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?

Zorodius

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

[tex]\lim_{x\rightarrow -1} \frac {2x^2-x-3}{x+1}[/tex]

[tex]= \lim_{x\rightarrow -1} \frac {(2x-3)(x+1)}{x+1}[/tex]

[tex]= \lim_{x\rightarrow -1} 2x-3[/tex]

= -5.

Caldus

You can use L'Hospital's rule:

[tex]\lim_{x\rightarrow -1} \frac {2x^2-x-3}{x+1}[/tex]

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

[tex]\lim_{x\rightarrow -1} \frac {2x^2-x-3}{x+1}[/tex]
= [tex]\lim_{x\rightarrow -1} \frac {4x-1}{1}[/tex]

Plug in and you get -5.

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ladyrae	How about this limit? Tweet 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?

Zorodius	I believe you want to factor the top, cancel out the hole in the graph, and then evaluate by substitution. [tex]\lim_{x\rightarrow -1} \frac {2x^2-x-3}{x+1}[/tex] [tex]= \lim_{x\rightarrow -1} \frac {(2x-3)(x+1)}{x+1}[/tex] [tex]= \lim_{x\rightarrow -1} 2x-3[/tex] = -5.

Caldus	You can use L'Hospital's rule: [tex]\lim_{x\rightarrow -1} \frac {2x^2-x-3}{x+1}[/tex] If you plug in, you get 0/0, so: [tex]\lim_{x\rightarrow -1} \frac {2x^2-x-3}{x+1}[/tex] = [tex]\lim_{x\rightarrow -1} \frac {4x-1}{1}[/tex] Plug in and you get -5.