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Math Help - Help with this limit?

  1. #1
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    Help with this limit?

    lim (1+2x) raised to the 3csc(2x) power
    x->0

    Sorry, I couldn't get it to display correctly but its (1+2x) ^ (3csc(2x)). I have it worked out by my professor, but am confused as to what he did. The first thing he did was take the natural log, and eventually the answer was e^3. How do you reach this answer?
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  2. #2
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by Mattpd View Post
    \lim_{x\rightarrow 0}(1+2x)^{3\csc(2x)}
    \lim_{x\rightarrow 0}\ (1+2x)^{3\csc(2x)} = \lim_{x\rightarrow 0}\ \exp(\ln{(1+2x)^{3\csc(2x)}}) = \lim_{x\rightarrow 0}\ \exp(3 \csc(2x)\ln{(1+2x)}) = \lim_{x\rightarrow 0}\ \exp(3\frac{\ln{(1+2x)}}{\sin(2x)}) = \lim_{x\rightarrow 0}\ \exp(3 \frac{\frac{2}{1+2x}}{2\cos(2x)}) =  e^{3}

    Where I used L'Hospitals in the second to last equality.
    Last edited by Anonymous1; April 6th 2010 at 10:05 AM.
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  3. #3
    Super Member Random Variable's Avatar
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     \lim_{x \to 0} \ (1+2x)^{3 \csc(2x)}

    let  y = \ln \Big((1+2x)^{3 \csc(2x)}\Big) = 3 \csc (2x) \ln (1+2x) = \frac{3 \ln(1+2x)}{\sin(2x)}

    then  \lim_{x \to 0} y = \lim_{x \to 0} \frac{3 \ln(1+2x)}{\sin(2x)}

    now use L'Hospital's rule

     = \lim_{x \to 0} \ \frac{3 \ \frac{2}{1+2x}}{2 \cos(2x)} = \frac{6}{2} = 3

    so  \lim_{x \to 0} \ (1+2x)^{3 \csc(2x)} = \lim_{x \to 0} e^{y} = e^{\lim_{x \to 0}y} = e^{3}
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  4. #4
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    Hello, Mattpd!

    This is a messy one . . .


    \lim_{x\to0}(1+2x)^{3\csc2x}

    Let: . y \:=\:(1+2x)^{3\csc2x}

    Take logs: . \ln(y) \;=\;\ln(1+2x)^{3\csc2x} \;=\;3\csc2x\cdot\ln(1+2x)

    We have: . \ln(y) \;=\;3\,\frac{\ln(1+2x)}{\sin2x} . . . which goes to \frac{0}{0}


    Apply L'Hopital: . 3\cdot\frac{\frac{1}{1+2x}\cdot2}{2\cos2x} \;=\;\frac{3}{(1+2x)\cos2x}

    Then: . \lim_{x\to0}\,\ln(y) \;=\;\lim_{x\to0}\frac{3}{(1+2x)\cos2x} \;=\;\frac{3}{1\cdot1} \;=\;3


    We have: . \lim_{x\to0}\,\ln(y) \:=\:3

    Therefore: . \lim_{x\to0} y \;=\;e^3

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  5. #5
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    L'Hospital help?

    Thanks all.
    Last edited by Mattpd; April 6th 2010 at 11:15 AM.
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