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Math Help - help on limit

  1. #1
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    help on limit

    1.) lim as theta-->pie/4 tanthetasin^2theta

    2.) lim x-->5 from the left 3x-15/|x-5|

    thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by badinmath View Post
    1.) lim as theta-->pie/4 tanthetasin^2theta

    2.) lim x-->5 from the left 3x-15/|x-5|

    thanks
    A. Use brackets to make you question clear, otherwise we are just guessing
    what you question realy is.

    B. I will assume you mean for your first question:

    lim as theta-->pi/4 tan[theta (sin(theta))^2)]

    Or:

    <br />
\lim_{\theta \to \pi /4} \tan [\theta ( sin( \theta))^2)]<br />

    Now as \tan [\theta ( sin( \theta))^2)] is continuouse at \pi/4 the limit may be evaluated by just plugging \pi/4 into this to get:

    <br />
\lim_{\theta \to \pi /4} \tan [\theta ( sin( \theta))^2)]=\tan [(\pi/4) ( sin( \pi/4))^2)] = \tan (\pi/8)

    C. By " lim x-->5 from the left 3x-15/|x-5|" I will assume you mean:

    lim x-->5 from the left (3x-15)/|x-5|

    or:

    <br />
\lim_{x \to 5 -} \frac{3x-15}{|x-5|}<br />

    <br />
\lim_{x \to 5 -} \frac{3x-15}{|x-5|} = 3 \frac{x-5}{|x-5|}<br />

    but when x<5, \frac{x-5}{|x-5|}=-1, so:

    <br />
\lim_{x \to 5 -} \frac{3x-15}{|x-5|} = 3 \frac{x-5}{|x-5|} = -3<br />

    RonL
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  3. #3
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    Hello, badinmath!

    The mathematical constant is \pi = "pi", not a dessert.


    If I read the first one correctly, it's ridiculous simple . . .

    1)\;\;\lim_{\theta\to\frac{\pi}{4}}\left[\tan\theta\!\cdot\sin^2\theta\right] \;=\;\tan\left(\frac{\pi}{4}\right)\!\cdot\sin^2\l  eft(\frac{\pi}{4}\right) \;=\;1\cdot\left(\frac{1}{\sqrt{2}}\right)^2 \;=\;1\cdot\frac{1}{2} \;=\;\frac{1}{2}

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  4. #4
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    hi

    thank you all. this forum is so helpful
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