Asked by assignment
limx-0 x-sinx/xcube
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Answered by
drwls
I assume that the numerator is (x - sinx)
In a Taylor series,
x - sinx = x^3/3! - x^5/5! + ...
Divide that by x^3
The limit as x->0 is therefore 1/3! = 1/6
In a Taylor series,
x - sinx = x^3/3! - x^5/5! + ...
Divide that by x^3
The limit as x->0 is therefore 1/3! = 1/6
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