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Math Help - Limits question

  1. #1
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    Limits question

    Any chance you can help me on this one mate?

    lim t -> 4 (4-t)/(5-[sqrt t^2 + 9])

    i subbed in x = 4 and solved and got 0. right?
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  2. #2
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    Re: Limits question

    Both the numerator and denominator goes to 0 as t goes to 4, so you can't just plug in t = 4. L'Hospitals rule works here though, so I suggest you use it.
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  3. #3
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    Re: Limits question

    When t=4 you don't get 0 you get 0/0 which isn't defined.
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Limits question

    Quote Originally Posted by Hooperoo View Post
    Any chance you can help me on this one mate?

    lim t -> 4 (4-t)/(5-[sqrt t^2 + 9])

    i subbed in x = 4 and solved and got 0. right?
    You can also do the following:
    \lim_{t \to 4} \frac{4-t}{5-\sqrt{t^2+9}} = \lim_{t \to 4} \frac{(4-t)(5+\sqrt{t^2+9})}{(5-\sqrt{t^2+9})(5+\sqrt{t^2+9})}
    =\lim_{t \to 4} \frac{(4-t)(5+\sqrt{t^2+9})}{25-(t^2+9)}
    =\lim_{t  \to 4} \frac{(4-t)(5+\sqrt{t^2+9})}{16-t^2}

    Proceed ...
    Last edited by Siron; March 24th 2012 at 05:12 AM.
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  5. #5
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    Re: Limits question

    Quote Originally Posted by Siron View Post
    You can also do the following:
    \lim_{t \to 4} \frac{4-t}{5-\sqrt{t^2+9}} = \lim_{t \to 4} \frac{(t-4)(5+\sqrt{t^2+9})}{(5-\sqrt{t^2+9})(5+\sqrt{t^2+9})}
    =\lim_{t \to 4} \frac{(t-4)(5+\sqrt{t^2+9})}{25-(t^2+9)}
    =\lim_{t  \to 4} \frac{(t-4)(5+\sqrt{t^2+9})}{16-t^2}

    Proceed ...
    Why has 4-t changed to t-4?
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: Limits question

    Quote Originally Posted by biffboy View Post
    Why has 4-t changed to t-4?
    It was a typo, thanks for noticing that!
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  7. #7
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    Re: Limits question

    Quote Originally Posted by Siron View Post
    You can also do the following:
    \lim_{t \to 4} \frac{4-t}{5-\sqrt{t^2+9}} = \lim_{t \to 4} \frac{(4-t)(5+\sqrt{t^2+9})}{(5-\sqrt{t^2+9})(5+\sqrt{t^2+9})}
    =\lim_{t \to 4} \frac{(4-t)(5+\sqrt{t^2+9})}{25-(t^2+9)}
    =\lim_{t  \to 4} \frac{(4-t)(5+\sqrt{t^2+9})}{16-t^2}

    Proceed ...
    Essentially, cant divide by 0, so limit is not defined?
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  8. #8
    MHF Contributor Siron's Avatar
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    Re: Limits question

    You can factore 16-t^2=(4-t)(4+t)
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  9. #9
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    Re: Limits question

    Quote Originally Posted by Siron View Post
    You can factore 16-t^2=(4-t)(4+t)
    so im getting lim t -> 4 (4-t)(5-sqrt t^2+9)/(4-t)(4+t)


    do i then just cancel the (4-t) and solve like that?
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  10. #10
    MHF Contributor Siron's Avatar
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    Re: Limits question

    Indeed.
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  11. #11
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Limits question

    $x_0=4$ is is called a removable discontinuity point.
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