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Tuff limit problem

  1. Jun 22, 2004 #1
    Okay basically i tried to facter out the top part of the rational function but it just doesnt seem corect.

    [tex] \lim{x\rightarrow2} [/tex]

    [tex] \frac{x^4-16}{x-2} [/tex]

    [tex] \lim{x\rightarrow2} [/tex]

    [tex] \frac{(x^2+4)(x^2-4}{X-2} [/tex]

    [tex] \lim{x\rightarrow2} \frac{(x+2)(x-2)(x^2-4)}{x-2} [/tex]

    [tex] \lim{x\rightarrow2} (x+2)(x^2-4)=0 [/tex]
    IS that correct, is it okay to factor out one and cancel?
    Last edited: Jun 22, 2004
  2. jcsd
  3. Jun 23, 2004 #2


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    Surely you mean

    [tex] \lim_{x\rightarrow2}~~\frac{(x^2-4)(x^2+4)}{x-2} [/tex]

    [tex]= \lim_{x\rightarrow2}~~ \frac{(x+2)(x-2)(x^2+4)}{x-2} [/tex]

    [tex] =\lim_{x\rightarrow2}~~ (x+2)(x^2+4)=32 [/tex]

    It is okay to factor out and cancel.
  4. Jun 23, 2004 #3
    yep, gokul is correct
  5. Jun 23, 2004 #4
    Hmm, sloppy hand writing on my paper seems to be the cause of confusion.
    after simplyfing [tex] (x^2-4) [/tex] I thought I had simplified [tex] (x^2+4) [/tex]...Even though this is a limit problem my gratatude is limitless... :rofl:
    yeah i konw its corny but i had to say it...thanks a million though Gokul43201.
  6. Jun 23, 2004 #5


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    The reason it is "correct to cancel" is that if f(x) and g(x) have the same value every where except at x= a, then [itex] \lim_{x\rightarrow a}f(x)= \lim_{x\rightarrow a}g(x)[/itex].

    In this problem, as long as x is not 2, [itex]\frac{x^4-16}{x-2}= (x^2+4)(x^2+2)[/itex] so the limits at 2 are the same.
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