calculus
posted by assignment on .
limx0 xsinx/xcube

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
posted by assignment on .
limx0 xsinx/xcube
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