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How about this limit?

  1. Jun 4, 2004 #1
    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. jcsd
  3. Jun 4, 2004 #2
    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.
     
  4. Jun 4, 2004 #3
    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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