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Math Help - find the limits of the function

  1. #1
    Junior Member
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    find the limits of the function

    consider the function

    g(x) = [((x-8)^2)*((x-1)^3)] / [sqrt(x-5)]

    i need to find the limits ..
    x -> 8+
    x-> 8-
    x-> 5+
    x-> 5-
    x-> 1+
    x-> 1-

    i've found
    x -> 8+ = infinity
    x-> 8- = infinity
    x-> 5+ = infinity
    x-> 5- = DNE
    x-> 1+ = DNE
    x-> 1- = DNE

    i thought that the last three didn't exsist because the square root was of a negative number when these were used, i know how to find limits but obviously i'm missing something here.
    i was almost certain that sqrt(-) cannot be done for this,
    please help me it's driving me mad

    thanks in advance,
    brittany
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  2. #2
    Super Member
    Joined
    Dec 2008
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    Scotland
    Posts
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    Quote Originally Posted by williamb View Post
    consider the function

    g(x) = [((x-8)^2)*((x-1)^3)] / [sqrt(x-5)]

    i need to find the limits ..
    x -> 8+
    x-> 8-
    x-> 5+
    x-> 5-
    x-> 1+
    x-> 1-

    i've found
    x -> 8+ = infinity
    x-> 8- = infinity
    x-> 5+ = infinity
    x-> 5- = DNE
    x-> 1+ = DNE
    x-> 1- = DNE

    i thought that the last three didn't exsist because the square root was of a negative number when these were used, i know how to find limits but obviously i'm missing something here.
    i was almost certain that sqrt(-) cannot be done for this,
    please help me it's driving me mad

    thanks in advance,
    brittany
    A limit exists for a function IFF  \displaystyle \lim_{x \to a^+} f(x) = \lim_{x \to a^-} f(x)

    So right away it doesn't make sense for the limit to exist for 3, but not 4.

    So let's look at it:

     \displaystyle \lim_{x \to a} \frac{(x-8)^2(x-1)^3}{\sqrt(x-5)}

     = \frac{ \bigg(\displaystyle \lim_{x \to a}(x-8)^2\bigg) \times \bigg(\displaystyle \lim_{x \to a} (x-1)^3 \bigg)}{ \displaystyle \lim_{x \to a} (\sqrt{x-5})}

     = \frac{ \bigg(\displaystyle \lim_{x \to a}(x-8)\bigg)^2 \times \bigg(\displaystyle \lim_{x \to a} (x-1) \bigg)^3}{ \sqrt{\displaystyle \lim_{x \to a}(x-5)}}

    Now for  a = 8^+ we get:

     = \frac{ (0)^2 \times (7)^3}{ \sqrt{(3)}} = 0

     a = 8^- we get:

     = \frac{ (0)^2 \times (7)^3}{ \sqrt{(3)}} = 0

     a = 5^+ we get:

     = \frac{ (-3)^2 \times (4)^3}{ \sqrt{(0)}} = \pm \infty

     a = 5^- is the one you had trouble with. Now you know that  5^- means it approaches 5 through negative values. But this doesn't always imply that we have a negative number. As x approaches 5 through negative values it must also pass through the interval  [0,5) , and these are non-negative. So we actually get:

     = \frac{ (-3)^2 \times (4)^3}{ \sqrt{(0)}} = \pm \infty

    And again, for  1^+ and  1^- you are getting negative square root, so indeed, an illegal operation.
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