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Math Help - Basic Limits Help

  1. #1
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    Basic Limits Help

    Reviewing for the final exam, forgot a couple simple rules.

    To solve a limit algebraically

    1. lim x -> -5 (x^2 +3x - 10) / (x + 5)

    If you plug in -5 it is 0/0 so I factored the top to (x+5)(x-2) but the bottom is still 0. I wrote no limit for the answer to this problem.

    2. lim x-> 3- (x)/(x^2 - 9)

    Again the limit is 0/0. I factored the bottom to be (x+3)(x-3) but plugging in 3 you wind up getting a number over 0 again. Again I think the answer is no limit.

    Am I forgetting something here?
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by KarlosK View Post
    Reviewing for the final exam, forgot a couple simple rules.

    To solve a limit algebraically

    1. lim x -> -5 (x^2 +3x - 10) / (x + 5)

    If you plug in -5 it is 0/0 so I factored the top to (x+5)(x-2) but the bottom is still 0. I wrote no limit for the answer to this problem.

    2. lim x-> 3- (x)/(x^2 - 9)

    Again the limit is 0/0. I factored the bottom to be (x+3)(x-3) but plugging in 3 you wind up getting a number over 0 again. Again I think the answer is no limit.

    Am I forgetting something here?
    1:

     \lim_{ x \to -5 } \frac{ x^2 +3x - 10 }{ x + 5 } = \lim_{ x \to -5 } \frac{ (x+ 5)(x-2) }{ x + 5 } = \lim_{ x \to -5 } (x-2) = -7

    2:

    \lim_{x \to 3^- } \frac{x}{x^2 - 9}

    We can do partial fractions here, but it is sufficient to note that 3 is being approached from the negative side (i.e. 2.99, 2.999, 2.9999, etc) and that makes the denominator a small negative quantity.

    Thus, a static number divided by an infinately small negative number is negative infinity.

    \lim_{x \to 3^- } \frac{x}{x^2 - 9} = - \infty
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