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Math Help - Limit Help

  1. #1
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    Limit Help

    I have no clue what to do here....please help

    lim as x goes to 0+ of [cos(pi/2-x)]^x

    and

    lim as x goes to 2+ of (1/x^2-4 - sqr(x-1)/(x^2-4)
    Last edited by poopforbrains; November 14th 2007 at 02:34 PM.
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  2. #2
    Eater of Worlds
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    Please use proper grouping symbols so we know exactly what you mean.

    And never spell 'pi' as 'pie'. This is a math forum, not a bakery.

    Anything to the 0 power is 1.

    \lim_{x\rightarrow{0^{+}}}[cos(\frac{\pi}{2}-x)]^{x}=1.
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  3. #3
    Math Engineering Student
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    Quote Originally Posted by galactus View Post
    \lim_{x\rightarrow{0^{+}}}[cos(\frac{\pi}{2}-x)]^{x}=1.
    Are you sure?
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  4. #4
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    Rewrite \frac{1}{x^{2}-4}-\frac{\sqrt{x-1}}{x^{2}-4}

    as \frac{1-\sqrt{x-1}}{x^{2}-4}

    Now, make the sub: u=\sqrt{x-1}

    You get:

    \lim_{u\rightarrow{1}}\frac{1-u}{u^{4}+2u^{2}-3}

    \lim_{u\rightarrow{1}}\frac{-1}{u^{3}+u^{2}+3u+3}=\boxed{\frac{-1}{8}}
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  5. #5
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    Quote Originally Posted by Krizalid View Post
    Are you sure?

    I checked it with my TI-92 and got this.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  6. #6
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    Hello, poopforbrains!

    I found a solution.
    . . I hope someone has a better way!


    \lim_{x\to0^+}\,\left[\cos\left(\frac{\pi}{2} - x\right)\right]^x

    Note that: . \cos\left(\frac{\pi}{2} - x\right) \;=\;\sin(x)

    So we have: . y \;=\;\left[\sin(x)\right]^x

    Take logs: . \ln(y) \;=\;\ln\left[\sin(x)\right]^x \;=\;x\cdot\ln[\sin(x)] \;=\;\frac{\ln[\sin(x)]}{\frac{1}{x}} . . . which approaches \frac{-\infty}{\infty}

    . . Apply L'Hopital: . \frac{\dfrac{\cos x}{\sin x}}{-\dfrac{1}{x^2}} \;=\;- \frac{x^2}{\tan x} . . . which approaches \frac{0}{0}

    . . Apply L'Hopital: . -\frac{2x}{\sec^2\!x} . . . which approaches -\frac{0}{1} \:=\:0

    Hence: . \lim_{x\to0^+} \ln(y)\;=\;0


    Therefore: . \lim_{x\to0^+} y \;=\;e^0 \;=\;1

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