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Math Help - Need a method for solving this..

  1. #1
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    Thumbs up [SOLVED]Need a method for solving this..

    \lim_{x \to 0} \frac{ln(1-x) + sin(x)}{1-cos^2(x)}

    Tried using L'Hopital and I know that's what you have to use but I still get stuck here..

    \lim_{x \to 0} \frac{\frac{1}{x-1}+cos(x)}{sin^2(x)}

    What am I doing wrong?
    Last edited by qleeq; February 10th 2011 at 10:44 PM.
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  2. #2
    MHF Contributor harish21's Avatar
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    You did not take the derivative of the denominator while using Lhopital's rule!
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  3. #3
    Junior Member guildmage's Avatar
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    Aren't you also supposed to take the derivative of the denominator?

    You just replaced 1-cos^2(x) with sin^2(x)
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  4. #4
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    Quote Originally Posted by guildmage View Post
    Aren't you also supposed to take the derivative of the denominator?

    You just replaced 1-cos^2(x) with sin^2(x)
    Right sorry..

    I meant

    \lim_{x \to 0} \frac{\frac{1}{x-1}+cos(x)}{-2cos(x)sin(x)}

    Even then I get stuck, can someone point me in the right direction?
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  5. #5
    MHF Contributor harish21's Avatar
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    \displaystyle \lim_{x \to 0} \dfrac{\ln(1-x)+sin(x)}{\sin^2(x)}.

    use Lhopitals rule to get:

    \displaystyle \lim_{x \to 0} \dfrac{\dfrac{1}{x-1}+\cos(x)}{2\sin(x)\cos(x)}



    \displaystyle =  \lim_{x \to 0} \dfrac{\dfrac{1+\cos(x)(x-1)}{x-1}}{\sin(2x)}



    \displaystyle = \lim_{x \to 0}\dfrac{1+(x-1)\cos(x)}{(x-1)\sin(2x)}


    use Lhopitals rule once again to get the answer.
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  6. #6
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    This goes to \displaystyle \frac{0}{0}, so you can use L'Hospital's Rule again. Also note that \displaystyle 2\sin{x}\cos{x} = \sin{2x}, which will make differentiating the denominator easier...
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  7. #7
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    Quote Originally Posted by harish21 View Post
    \displaystyle \lim_{x \to 0} \dfrac{\ln(1-x)+sin(x)}{\sin^2(x)}.

    use Lhopitals rule to get:

    \displaystyle \lim_{x \to 0} \dfrac{\dfrac{1}{x-1}+\cos(x)}{2\sin(x)\cos(x)}





    \displaystyle  \lim_{x \to 0} \dfrac{\dfrac{1+\cos(x)(x-1)}{x-1}}{\sin(2x)}



    \displaystyle \lim_{x \to 0}\dfrac{1+(x-1)\cos(x)}{(x-1)\sin(2x)}


    use Lhopitals rule once again to get the answer.
    \displaymode = \dfrac{(x-1)(-\sin(x))+ \cos(x)(1)}{(x-1)cos(2x)(2)+sin(2x)(1)}


    \displaymode = \dfrac{-\sin(x)+\sin(x)+\cos(x)}{2x\cos(x)-2\cos(x)+\sin(2x)}

    Then plugging it all in..
    -x\sin(0)+\sin(0)+\cos(0) = 0 + 0 + 1 = 1

    2(0)\cos(0)-2\cos(0)+\sin(2(0)) = 0 - 2(1) + 0 = -2

    \displaymode = -\frac{1}{2}

    Have I made any errors?
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  8. #8
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by qleeq View Post
    \displaymode = \dfrac{-\sin(x)+\sin(x)+\cos(x)}{2x\cos(x)-2\cos(x)+\sin(2x)}

    Have I made any errors?
    The denominator should be 2x\cos(2x)-2\cos(2x)+\sin(2x). You missed typing the 2s I guess.

    Otherwise there are no serious errors. good job!
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  9. #9
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    Quote Originally Posted by harish21 View Post
    The denominator should be 2x\cos(2x)-2\cos(2x)+\sin(2x). You missed typing the 2s I guess.

    Otherwise there are no serious errors. good job!
    Must have accidentally removed them while I was editing.. Anyway, thanks a lot for your help. I had one last question though.

    \displaystyle  \lim_{x \to 0} \dfrac{\dfrac{1+\cos(x)(x-1)}{x-1}}{\sin(2x)}


    \displaystyle \lim_{x \to 0}\dfrac{1+(x-1)\cos(x)}{(x-1)\sin(2x)}

    Can you explain to me how he arrived at the firstequation from \displaystyle \lim_{x \to 0} \dfrac{\dfrac{1}{x-1}+\cos(x)}{2\sin(x)\cos(x)}?

    I understand the second part, but the first part has me confused.

    How did he get \dfrac{\dfrac{1+cos(x)(x-1)}{(x-1)}}{sin(2x)} from \dfrac{\dfrac{1}{x-1}+cos(x)}{2sin(x)cos(x)} ?

    I understand the sin(2x) = 2sin(x)cos(x) part.
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  10. #10
    MHF Contributor harish21's Avatar
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    \dfrac{1}{x-1}+\cos(x)

    = \dfrac{1}{x-1}+ \bigg(\dfrac{\cos(x)}{1} \times \dfrac{x-1}{x-1}\bigg)

     = \dfrac{1+\cos(x)(x-1)}{x-1}
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  11. #11
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    Quote Originally Posted by harish21 View Post
    \dfrac{1}{x-1}+\cos(x)

    = \dfrac{1}{x-1}+ \bigg(\cos(x) \times \dfrac{x-1}{x-1}\bigg)

     = \dfrac{1+\cos(x)(x-1)}{x-1}
    You are amazing. Thanks
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  12. #12
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    Actually, in this case it's easier to use L'Hospital's Rule the second time BEFORE doing any algebraic manipulation...

    \displaystyle \lim_{x \to 0}\frac{\frac{1}{x-1} + \cos{x}}{-2\sin{x}\cos{x}} = \lim_{x \to 0}\frac{(x-1)^{-1} + \cos{x}}{-\sin{2x}}

    \displaystyle = \lim_{x \to 0}\frac{-(x-1)^{-2} - \sin{x}}{-2\cos{2x}} by L'Hospital's Rule.

    The limit is now evaluated through direct substitution.
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