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Math Help - Finding the limit

  1. #1
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    Finding the limit

    Question is find the value of
    [tex]
    \mathop {\lim }\limits_{x \to 9} \frac{{t - 9}}{{3 - \sqrt t }}
    /MATH]

    This is as far as I get

    <br />
 \mathop {\lim }\limits_{x \to 9} \frac{{t - 9}}{{3 - \sqrt t }} \\ <br />
  = \mathop {\lim }\limits_{x \to 9} \frac{{t - 9}}{{3 - \sqrt t }} \times \frac{{3 + \sqrt t }}{{3 + \sqrt t }} \\ <br />
  = \mathop {\lim }\limits_{x \to 9} \frac{{t^{\frac{3}{2}}  + 3t - 9\sqrt t  - 27}}{{9 - t}} \\ <br />

    At this stage I'm left with a divide by zero case, if I sub in 9 in an attempt to find the limit. Where do I go from here?
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  2. #2
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    <br />
\mathop {\lim }\limits_{t \to 9} \frac{{t - 9}}{{3 - \sqrt t }}<br />

    Since you end up with a 0/0 indeterminate case, you can use L'Hospital's rule to solve the limit. That is, if you are given

    <br />
\mathop {\lim }\limits_{x \to a} \frac{f(x)}{g(x)}<br />

    and you end up with 0/0, the limit is

    <br />
\mathop {\lim }\limits_{x \to a} \frac{f'(x)}{g'(x)}<br />

    So take the derivative of the numerator, then the derivative of the denominator, then solve for the limit:

    <br />
\mathop {\lim }\limits_{t \to 9} \frac{1}{- 0.5t^{-0.5}} = \mathop {\lim }\limits_{t \to 9} -2 \sqrt {t} = -6 <br />
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  3. #3
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    You could also not multiply out the numerator:
    \lim_{x\to{9}}\frac{t-9}{3-\sqrt{t}}=\lim_{x\to{9}}\frac{t-9}{3-\sqrt{t}}\cdot\frac{3+\sqrt{t}}{3+\sqrt{t}}=\lim_{  x \to 9} \frac{(t-9)(3+\sqrt{t})}{9 - t}
    And since t-9 and 9-t are opposites, their quotient is -1, and we get that the above
    \lim_{x\to{9}}\frac{t-9}{3-\sqrt{t}}=\lim_{x \to 9}-(3+\sqrt{t})=-6.

    --Kevin C.
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  4. #4
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    Quote Originally Posted by TwistedOne151 View Post
    You could also not multiply out the numerator:
    \lim_{x\to{9}}\frac{t-9}{3-\sqrt{t}}=\lim_{x\to{9}}\frac{t-9}{3-\sqrt{t}}\cdot\frac{3+\sqrt{t}}{3+\sqrt{t}}=\lim_{  x \to 9} \frac{(t-9)(3+\sqrt{t})}{9 - t}
    And since t-9 and 9-t are opposites, their quotient is -1, and we get that the above
    \lim_{x\to{9}}\frac{t-9}{3-\sqrt{t}}=\lim_{x \to 9}-(3+\sqrt{t})=-6.

    --Kevin C.
    I think if we note that (t-9)=-(3+\sqrt{t})(3-\sqrt{t}) then it is easier.
    Last edited by mr fantastic; January 6th 2009 at 02:18 AM. Reason: Clarified the language
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  5. #5
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    Thanks Kevin. I've seen that done before but I couldn't remember what to do thanks.
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  6. #6
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    Hello, Craka!

    Here's another approach . . .


    \lim_{t\to9}\, \frac{t - 9}{3 - \sqrt{t}}

    Factor -1 out of the denominator: . \lim_{t\to9}\,\frac{t-9}{-(\sqrt{t} - 3)}

    Factor the numerator: . \lim_{t\to9}\,-\frac{(\sqrt{t} - 3)(\sqrt{t} + 3)}{\sqrt{t}-3}

    Reduce: . \lim_{t\to9}\,-(\sqrt{t} + 3)

    Evaluate: . -(\sqrt{9} + 3) \;=\;-6

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