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Math Help - confusing limit..

  1. #1
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    confusing limit..

    I have the limit as x goes to 0 ln(cotx)/(e^csc^2(x))
    it reads as the natural log of cotx divided by e raised to the csc squared x
    the csc is csc^2(x). cotx as x goes to 0 is 1/0, which is undefined. How would I start this?
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  2. #2
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    \lim_{x\rightarrow{0}}\frac{ln(cot(x))}{e^{csc^{2}  (x)}}

    Try letting u=e^{csc^{2}(x)}

    csc^{2}(x)=cot^{2}(x)+1

    u=e^{cot^{2}(x)+1}

    ln(u)=cot^{2}(x)+1

    \sqrt{ln(u)-1}=cot(x)

    ln(\sqrt{ln(u)-1})=\frac{ln(ln(u)-1)}{2}

    Then you get:

    \frac{1}{2}\lim_{u\rightarrow{\infty}}\frac{ln(ln(  u)-1)}{u}

    Now use L'Hopital:

    \frac{1}{2}\lim_{u\rightarrow{\infty}}\frac{1}{u(l  n(u)-1)}=0
    Last edited by galactus; July 14th 2007 at 02:52 PM.
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  3. #3
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    Quote Originally Posted by galactus View Post
    \lim_{x\rightarrow{0}}\frac{ln(cot(x))}{e^{csc^{2}  (x)}}

    Try letting u=e^{csc^{2}(x)}

    Then you get:

    \frac{1}{2}\lim_{u\rightarrow{\infty}}\frac{ln(ln(  u)-1)}{u}

    Now use L'Hopital:

    \frac{1}{2}\lim_{u\rightarrow{\infty}}\frac{1}{u(l  n(u)-1)}=0
    I saw you changed x goes to 0 to u goes to infinity. Why does u not go to 0?
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  4. #4
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    Because we done a change of variables. Just like when you sub in integration, you have to change the limits.

    \lim_{x\rightarrow{0}}e^{csc^{2}(x)}={\infty}
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