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Math Help - Improper Integrals My Tutor Cannot Understand!!!

  1. #1
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    Improper Integrals My Tutor Cannot Understand!!!

    Here are some problems I am having a hard time with. If anyone knows how to solve them It would be amazing. My tutor cannot even figure these out!!!

    1. limx->2- sqrt(4-x^2)/x-2

    2. limx->0+ x^3cotx

    3. limx->infinity (1+1/x)^x

    4. limx->2+ (1/x^2-4)-(sqrtx-1/x^2-4)
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  2. #2
    Super Member wingless's Avatar
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    Are these limits or integrals? If they are integrals, what are their bounds?
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  3. #3
    Senior Member polymerase's Avatar
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    Do you mean something like this:

    \int ^2_k \frac{\sqrt{4-x^2}}{x-2} where k is a constant  -2\le k<2

    If this is what you mean then to solve this you have to treat it like a limit as such:

    \int ^2_k \frac{\sqrt{4-x^2}}{x-2}=\displaystyle \lim_{t\to 2^{-}} \int ^t_k \frac{\sqrt{4-x^2}}{x-2}. where t is any constant k<t<2. Then you would do your regular integration by letting x=2 \sin u and then plugging in t and k into your F(x). You end up with:

    \lim_{t\to 2^{-}} \left(\sqrt{4-t^2}-2\arcsin \left(\frac{t}{2}\right)\right) (I have omitted the part where you plug in k since those will just be numbers thus not affected by the limit).

    You solve the limit, you end up with -\pi.

    Therefore, \int ^2_k \frac{\sqrt{4-x^2}}{x-2}=-\pi-\left(\sqrt{4-k^2}-2\arcsin \left(\frac{k}{2}\right)\right)
    Last edited by polymerase; March 15th 2008 at 04:11 PM.
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  4. #4
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    Perhaps you should recognize the third one as a familiar e limit.

    \lim_{x\rightarrow{\infty}}(1+\frac{1}{x})^{x}=e

    Let t=\frac{1}{x}, \;\ x=\frac{1}{t}

    Then we get another fundamental limit:

    \lim_{t\rightarrow{0}}(1+t)^{\frac{1}{t}}=e

    To prove the latter, we can build on the differentiability of ln(x) at x=1.

    \frac{d}{dx}[ln(x)]=\frac{1}{x}=1

    1=\lim_{t\rightarrow{0}}\frac{ln(1+t)-ln(1)}{t}

    =\lim_{t\rightarrow{0}}\frac{ln(1+t)}{t}

    =\lim_{t\rightarrow{0}}ln(1+t)^{\frac{1}{t}}

    And it follows:

    e=e^{\lim_{t\rightarrow{0}}ln(1+t)^{\frac{1}{t}}}

    e=\lim_{t\rightarrow{0}}e^{ln(1+t)^{\frac{1}{t}}}

    =\lim_{t\rightarrow{0}}(1+t)^{\frac{1}{t}}
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  5. #5
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    Quote Originally Posted by daciernoj View Post
    Here are some problems I am having a hard time with. If anyone knows how to solve them It would be amazing. My tutor cannot even figure these out!!!

    1. limx->2- sqrt(4-x^2)/x-2

    2. limx->0+ x^3cotx

    3. limx->infinity (1+1/x)^x

    4. limx->2+ (1/x^2-4)-(sqrtx-1/x^2-4)
    4. is ambiguous and needs more brackets or latex typesetting.

    -----------------------------------------------------------------------------------------------

    1. = \lim_{x \rightarrow 2^{-}} \frac{\sqrt{(2-x)(2+x)}}{-(2-x)} = - \lim_{x \rightarrow 2^{-}}\frac{\sqrt{2+x}}{\sqrt{2-x}} = - \frac{4}{0^{+}} = -\infty.

    Those who are easily offended, please turn a blind eye to the second last bit

    ------------------------------------------------------------------------------------------------

    2. = \lim_{x \rightarrow 0^{+}} \frac{x^3 \, \cos x}{\sin x} = \lim_{x \rightarrow 0^{+}} \frac{x^3 \, \cos x}{x - \frac{x^3}{3!} + ....} = \lim_{x \rightarrow 0^{+}} \frac{x^2 \, \cos x}{1 - \frac{x^2}{3!} + ....} = you should be able to take it from here.
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