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Math Help - A Limit (Maybe using the infinite series?)

  1. #1
    Ted
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    A Limit (Maybe using the infinite series?)

    Hello.
    I stopped at this one:

    \lim_{ x \to 0} \dfrac{1-\frac{1}{2}x^2-cos\left( \frac{x}{1-x^2}\right)}{x^4}

    I did not try the L'Hospital's Rule, Am searching for a perfect solution.
    Maybe using the infinite series ?
    But it seems brutal.


    Any help?
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  2. #2
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    transform cos into sin using \cos{2x}=1-2\sin^2{x} then u can apply (sin x)/x -> 1 limit... then see what you have left
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  3. #3
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     \lim_{x\to 0 } \frac{ 1 - \frac{x^2}{2} - \cos( \frac{x}{1-x^2} ) }{x^4}

     = - \frac{1}{24} +  \lim_{x\to 0 }  \frac{ 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cos( \frac{x}{1-x^2} ) }{x^4 }

     =  - \frac{1}{24} +  \lim_{x\to 0 }  \frac{ \cos(x) +  O(x^6) - \cos( \frac{x}{1-x^2} ) }{x^4 }

     =  - \frac{1}{24} +  \lim_{x\to 0 }  \frac{ 2 \sin\left( \frac{x^3}{2(1-x^2)} \right) \sin\left( \frac{x(2-x^2)}{1-x^2} \right) }{ x^4 }

     =  - \frac{1}{24} +  2\lim_{x\to 0 }  \frac{ \sin\left( \frac{x^3}{2(1-x^2)} \right) }{x^3} \lim_{x\to 0 }  \frac{\sin\left( \frac{x(2-x^2)}{2(1-x^2)} \right) }{ x }

     =  - \frac{1}{24} +  2 ~ (\frac{1}{2}) ~( 1 ) = 1 - \frac{1}{24} = \frac{23}{24}
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  4. #4
    Ted
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    Utherr :
    Thanks. I will try it.

    simplependulum:
    Thanks. But sorry, I do not know what does the symbol O mean.
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  5. #5
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    double post :-s
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  6. #6
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    Quote Originally Posted by Ted View Post
    Utherr :
    Thanks. I will try it.

    simplependulum:
    Thanks. But sorry, I do not know what does the symbol O mean.
    it doesn't matter , it means the order of x is 6 so O(x^6) tends to zero when x does too (6 > 4 ) .
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