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Limit of Lim h->0 ((2+h^2) - 8)/h

  1. Sep 19, 2012 #1
    I saw my professor solve for the limit of ((2+h)2 - 8)/h as h approaches 0 and it was defined, but for the life of me I can't remember how. Everyone I have asked can't seem to figure it out either, so I was hoping someone here could explain.
     
  2. jcsd
  3. Sep 19, 2012 #2

    pwsnafu

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    I'm going to assume the thread title is a typo.

    Hint: let ##h = 0.1, 0.01, 0.001, \ldots## and manually calculate the expression for each ##h##.
     
  4. Sep 19, 2012 #3

    You wrote

    $$\frac{(2+h)^2-8}{h}$$

    and the limit of this thing when [itex]\,h\to 0\,[/itex] doesn't exist. But perhaps you meant

    $$\lim_{h\to 0}\frac{(2+h)^3-8}{h}=\lim_{h\to 0}\frac{8+12h+6h^2+h^3-8}{h}=\lim_{h\to 0}\;(12+6h+h^2)=12$$

    DonAntonio
     
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