Question
limx-0 x-sinx/xcube
Answers
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