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?

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 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.
 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.