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Different Way to Solve a Limit of an Indeterminate Equation

  1. Aug 14, 2008 #1
    lim_{x\rightarrow-2}\frac{x^3 + 8}{x+2}

    Ok, I have solved this using synthetic division in which the limit was 12, however I was wondering if there was a way to solve this without using synthetic division (seeing as I hate the process of it).

    I've solved plenty of other indeterminate equations by factoring out a polynomial and then being able to cancel out so that I can then plug in x, but I wasn't able to do so with this one.

    I'm just curious if there are any other ways to solve this specific limit.

  2. jcsd
  3. Aug 14, 2008 #2
    L'hospital rule (applies only when it is in forms like 0/0, inf/inf)

    differentiate both num and den, and then sub in your -2
  4. Aug 14, 2008 #3
    Yeah I started reading up on that rule, but didn't understand it fully until your reply. So the derivative of the numerator is 3x^2 and the denominator is 1 so then just sub in the -2 and you get 12. Got it, thanks a lot for clearing up that rule for me.
  5. Aug 14, 2008 #4


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    Not that this is much different than using synthetic division, but the numerator is a sum of two cubes, which factors according to

    [tex]a^3 + b^3 = (a + b) \cdot (a^2 - ab + b^2)[/tex] . So, in fact, this numerator can be factored and the (x + 2) term can be cancelled.
    Last edited: Aug 14, 2008
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