lim as x -> 0 sin(2x)/x^3 - Wolfram|Alpha
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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?
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.
lim as x -> 0 sin(2x)/x^3 - Wolfram|Alpha
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I'm pretty comfortable with 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.