Using the Maclaurin series for sin(x)=x−x33!+x55!−⋯\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdotssin(x)=x−3!x3+5!x5−⋯, find limx→0sin(x)−xx3\displaystyle \lim_{x \to 0} \frac{\sin(x) - x}{x^3}x→0limx3sin(x)−x.
000
−16-\frac{1}{6}−61
16\frac{1}{6}61
The limit does not exist