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Math Help - Limits

  1. #1
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    Last edited by gamboo; October 27th 2009 at 07:47 PM.
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  2. #2
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    3(b): \lim_{t\to 5} \frac{t^2+3t-40}{t^2-25} = \frac{0}{0}

    Derive the top and bottom and resubstitute t = 5.

    3(c): Same as above

    I didn't look at 3a because I'm too tired for Trig, but using L'Hospitals Rule makes them rather simple. If you can't do the derivatives, let me know.
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  3. #3
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    Quote Originally Posted by Open that Hampster! View Post
    3(b): \lim_{t\to 5} \frac{t^2+3t-40}{t^2-25} = \frac{0}{0}

    Derive the top and bottom and resubstitute t = 5.

    3(c): Same as above

    I didn't look at 3a because I'm too tired for Trig, but using L'Hospitals Rule makes them rather simple. If you can't do the derivatives, let me know.
    I can't, sorry.
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  4. #4
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    \frac{sin(4-4cos(2t))}{1-cos(2t)} = \frac{sin(4-4cos(2t))}{1-cos(2t)} \cdot \frac{4}{4} = 4\cdot \frac{sin(4-4cos(2t))}{4-4cos(2t)}

    Now, using the fact that \lim_{u(x) \to 0} \frac{sin(u(x))}{u(x)} = 1 and \lim_{t \to 0} 4-4cos(2t) = 0, we get:

    \lim_{t \to 0} \frac{sin(4-4cos(2t))}{1-cos(2t)} = \lim_{t \to 0} 4\cdot \frac{sin(4-4cos(2t))}{4-4cos(2t)} = 4 \cdot 1 = 4

    Also, a simpler approach to (b) would be to note that t^2-25 = (t+5)(t-5) and that t^2+3x-40 = (t-5)(t+8), therefore:

    \lim_{t \to 5} \frac{t^2+3x-40}{t^2-25} = \lim_{t \to 5} \frac{(t+8)(t-5)}{(t+5)(t-5)} = \lim_{t \to 5} \frac{t+8}{t+5} = \frac{13}{10} as expected.
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