Determine the values of aaa and bbb such that limx→0asinx+bsin(2x)x3=1\lim_{x \to 0} \frac{a \sin x + b \sin(2x)}{x^3} = 1limx→0x3asinx+bsin(2x)=1.
a=−4,b=1a=-4, b=1a=−4,b=1
a=4,b=−1a=4, b=-1a=4,b=−1
a=−2,b=1a=-2, b=1a=−2,b=1
a=2,b=−1a=2, b=-1a=2,b=−1