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Math Help - lost of issues

  1. #1
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    lost of issues

    I meant lots of issues, sorry

    The problem says to : Evaluate the improper integrals. Show the proper use of limits.

    The integral of 1/(x^2-9) from 3 to infinity.

    I can get it into the limit form and solve the integral part but I can't solve the limit. It works out to the limit as a approaches 3 from the left(+) of (1/9 ln 1/4 - 1/9 ln |(a-3)/(a+3)|) + limit as b approaches infinity from the right (-) of (1/9 ln |(b-3)/(b+3)| - 1/9 ln 1/4) How do you solve that?

    Also I can't figure out that the limit of e^(-1/x) as x approaches 0 from the right and from the left is.

    Finially I don't know where to even start on the integral of
    sqrt(9-(lnx)^2)
    (lnx^x)
    Last edited by Amberosia32; March 16th 2010 at 03:59 PM. Reason: mistake
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  2. #2
    Super Member Anonymous1's Avatar
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    \int_{3}^{\infty} \frac{1}{(x^2-9)} = -1/3 \tanh^{-1} (x/3).

    \lim_{x\rightarrow \infty} -1/3 \tanh^{-1} (x/3) + \lim_{x\rightarrow 3}1/3 \tanh^{-1} (x/3)

    Now just use the definition \tanh^{-1}(x) = 1/2[\ln(1+x) - \ln(1-x)].

    \lim_{x \rightarrow 0} e^{(-1/x)} = \lim_{x \rightarrow 0} \frac{1}{e^{(1/x)}} = \frac{1}{\infty} = 0
    Last edited by Anonymous1; March 16th 2010 at 03:35 PM.
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  3. #3
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    Quote Originally Posted by Anonymous1 View Post
    \int_{2}^{\infty} \frac{1}{(x^2-9)} = -1/3 tanh^{-1} (x/3).

    I assume you want \lim_{x\rightarrow +-3} -1/3 tanh^{-1} (x/3)?
    I typed it wrong the lower number should have been 3 not 2. I'm sorry.
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  4. #4
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    I solved the integral part to be 1/9 ln |(x-3)/(x+3)|
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  5. #5
    Super Member Anonymous1's Avatar
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    So what are you taking the limit of?
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  6. #6
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    Quote Originally Posted by Anonymous1 View Post
    So what are you taking the limit of?
    I hope this makes sense. I don't know how to type it.

    ∫ 1/(x^2-9) from 3 to infinity, ok so the 3 and the infinity make it improper, so it becomes:
    limit as a(bottom number) approaches 3 of ∫ 1/(x^2-9) w/ the bottom number being 3 and the top number being any number you choose, so let's say 5 from the left(+). + limit as b(top number approaches infinity of ∫ 1/(x^2-4) with the top number being infinity and the bottom number being the 5 I picked from the right (-).

    So it is the limit as a approaches 3 from the left(+) of (1/9 ln |(5-3)/(5+3)| - 1/9 ln |(a-3)/(a+3)|) + limit as b approaches infinity from the right (-) of (1/9 ln |(b-3)/(b+3)| - 1/9 ln |(5-3)/(5+3)|), which is

    the limit as a approaches 3 from the left(+) of (1/9 ln 1/4 - 1/9 ln |(a-3)/(a+3)|) + limit as b approaches infinity from the right (-) of (1/9 ln |(b-3)/(b+3)| - 1/9 ln 1/4)

    After that I'm lost
    Last edited by Amberosia32; March 16th 2010 at 03:52 PM.
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