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Math Help - Another limit question

  1. #1
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    Another limit question

    lim as h -> 0 (6 + h) to the power of 3 subtract 216 divided by h

    Any help?
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  2. #2
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    You can expand (6+h)^3 or take advantage of property a^3-b^3=(a-b)(a^2+ab+b^2) on numerator.
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  3. #3
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    Quote Originally Posted by Hockey_Guy14 View Post
    lim as h -> 0 (6 + h) to the power of 3 subtract 216 divided by h

    Any help?
    This is apparently
    \lim_{h \to 0} \frac{(6 + h)^3 - 216}{h}

    Expand the cubic:
    = \lim_{h \to 0} \frac{(216 + 108h + 18h^2 + h^3)  - 216}{h}

    Now simplify:
    = \lim_{h \to 0} \frac{108h + 18h^2 + h^3}{h}

    Factor:
    = \lim_{h \to 0} \frac{h(108 + 18h + h^2)}{h}

    Cancel:
    = \lim_{h \to 0} (108 + 18h + h^2)

    And finally take the limit:
    = 108

    Many limits in the "secant form" above can be done by this procedure.

    -Dan
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