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Math Help - Solving for unknowns in a limit problem.

  1. #1
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    Solving for unknowns in a limit problem.

    \lim_{x\to 0}\frac{(sin 2x)}{(x^3)}+a+(\frac{b}{x^3})=0

    The problem states "find a and b such that the limit as x approaches 0 is equal to 0."

    My strategy has been to use L'Hospital's Rule to evaluate the first term, ((sin 2x)/x^3), and then find an a and b that cancel out the leading term near x=0.

    OK so my first problem is evaluating the leading term as x approaches 0. Since sine is positive between 0 and 2 and x cubed is positive , for negative x the quotient is positive as well, then the limit of the first term as x approaches 0 has to be positive infinity. I am at a loss as to how I should go about showing that this is true. (applying L'Hospital's leads to 2/0 so that doesn't seem to be too fruitful. I have tried substituting [2 cos x sin x])

    The second difficulty is finding an a and b that cancel out the leading term. I think I need a and b such that their limit as x approaches 0 is not only negative infinity, but they have to decrease at the same rate that the leading term increases.

    Perhaps?

    \frac{-(sin 2x)}{(x^3)}= \frac{-1*(sin 2x)}{x^3} + \frac{-1}{x^3}

    a = \frac{-1*(sin 2x)}{x^3}
    b = -1

    This seems really stupid though...


    I'll be trying to figure out latex over the next 15min or so. Just looking to clean up the problem statement.
    Last edited by bkbowser; March 24th 2012 at 08:34 AM. Reason: Conversion into latex. Also I fixed a negative sign underneath "Perhaps?".
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  2. #2
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    Re: Solving for unknowns in a limit problem.

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  3. #3
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    Re: Solving for unknowns in a limit problem.

    I'm pretty comfortable with \lim {x \to 0}\frac{sin 2x}{x^3}=0 now.

    I don't really care for the rest of my solution though. All I did was set the unknowns into the additive inverse of the known. I just made the function, 0. Rather I'd prefer a solution that makes the limit as x approaches 0, 0. While the rest of the function is still something non-constant.
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