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Math Help - Limits

  1. #1
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    Limits

    Having trouble on these two questions. I have no idea and help would be much appreciated. Thanks.

    Find
     \lim_{t \to 3} \frac{t^2 - 2t - 3}{t^2 + 4t - 21}

    Find
     \lim_{t \to \infty} \sqrt{t^2 + t} - t
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by WWTL@WHL View Post
    Having trouble on these two questions. I have no idea and help would be much appreciated. Thanks.

    Find
     \lim_{t \to 3} \frac{t^2 - 2t - 3}{t^2 + 4t - 21}
    Factorise !
    t^2-2t-3=(t-3)(t+1)

    and t^2+4t-21=(t-7)(t+3)

    Find
     \lim_{t \to \infty} \sqrt{t^2 + t} - t
    Multiply and divide by its conjugate, that is \frac{\sqrt{t^2+t}{\color{red}+}t}{\sqrt{t^2+t}+t}
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  3. #3
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    Thanks for the quick reply, Moo.

    Quote Originally Posted by Moo View Post
    Hello,

    Factorise !
    t^2-2t-3=(t-3)(t+1)

    and t^2+4t-21=(t-7)(t+3)
    Sorry, I'm still have trouble with this. So substituting 3 in for t, and you have 0 on the numerator, and -3 on the denominator, which suggests it'll tend to 0, as t approaches 3?

    But t^2+4t-21 = 0 when t=3 as well. So then you'd have 0/0 which is undefined?


    Multiply and divide by its conjugate, that is \frac{\sqrt{t^2+t}{\color{red}+}t}{\sqrt{t^2+t}+t}
    Excellent. Thanks a lot. It tends to 1/2.
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  4. #4
    Moo
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    Quote Originally Posted by WWTL@WHL View Post
    Sorry, I'm still have trouble with this. So substituting 3 in for t, and you have 0 on the numerator, and -3 on the denominator, which suggests it'll tend to 0, as t approaches 3?

    But t^2+4t-21 = 0 when t=3 as well. So then you'd have 0/0 which is undefined?
    Excuse me

    It's actually t^2+4t-21=(t-3)(t+7)

    Do you see any simplification ?
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  5. #5
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    Thanks.

    <br />
\lim_{t \to 3} \frac{t^2 - 2t - 3}{t^2 + 4t - 21}<br />
= \lim_{t \to 3} \frac{t+1}{t+7} = \frac{2}{5}

    Does that seem ok?
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  6. #6
    Moo
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    Quote Originally Posted by WWTL@WHL View Post
    Thanks.

    <br />
\lim_{t \to 3} \frac{t^2 - 2t - 3}{t^2 + 4t - 21}<br />
= \lim_{t \to 3} \frac{t+1}{t+7} = \frac{2}{5}

    Does that seem ok?
    Yes

    You should add a step :
    <br />
\lim_{t \to 3} \frac{t^2 - 2t - 3}{t^2 + 4t - 21}=\lim_{t \to 3} \frac{(t-3)(t+1)}{(t-3)(t+7)}=\ldots

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