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Thread: limit 2

  1. #1
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    limit 2

    lim 1-cos2x
    x-->0 -------------=?
    x^2+x.sinx

    I applied L'Hospital but couldn't get rid of x.
    I tried another way

    1-cos2x=1-(cos^x-sin^x)=sin^2x

    sin^2x
    ----------
    x^2+xsinx

    I tried to divide both sides by 2x to apply sandwich rule but I couldn't get anything.
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  2. #2
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    Re: limit 2

    Quote Originally Posted by kastamonu View Post
    lim 1-cos2x
    x-->0 -------------=?
    x^2+x.sinx

    I applied L'Hospital but couldn't get rid of x.
    I applied L'Hospital's rule twice, and that was enough. It works out that way if you persevere.
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  3. #3
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    Re: limit 2

    2cosx/2x+sinx+xcosx I got.

    And found the answer as 1/2 but book says 1. I made a mistake somewhere.
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  4. #4
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    Re: limit 2

    Quote Originally Posted by kastamonu View Post
    2cosx/2x+sinx+xcosx <----- You need grouping symbols around the denominator (whether or not this expression is correct)
    to show what you intend.


    I got.

    And found the answer as 1/2 but book says 1. I made a mistake somewhere.
    Realize that there is supposed to be limit notation in front of each of the following lines that have expressions:


    \dfrac{1 - cos(2x)}{x^2 + xsin(x)}


    L'Hospital's rule (with the limit understood being applied):


    \dfrac{sin(2x)*2}{2x + sin(x) + xcos(x)}


    \dfrac{2sin(2x)}{2x + sin(x) + xcos(x)}


    L'Hospital's rule (with the limit understood being applied) a second time:


    \dfrac{2cos(2x)*2}{2 + cos(x) + cos(x) - xsin(x)}


    \dfrac{4cos(2x)}{2 + 2cos(x) - xsin(x)}


    Now, take the limit as x approaches 0 of the last expression.
    Last edited by greg1313; May 31st 2017 at 10:44 AM.
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  5. #5
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    Re: limit 2

    Many Thanks.

    May I also use Sandwich rule after applying L'Hospital? I am asking generally.
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