limh→0sin(π3+h)−sin(π3)h=\displaystyle \lim_{h \to 0} \frac{\sin\left(\frac{\pi}{3}+h\right) - \sin\left(\frac{\pi}{3}\right)}{h} =h→0limhsin(3π+h)−sin(3π)=A000B12\frac{1}{2}21check_circleC111D32\frac{\sqrt{3}}{2}23ExplanationThis is f′(π/3)f'(\pi/3)f′(π/3) where f(x)=sinxf(x)=\sin xf(x)=sinx, so cos(π/3)=1/2\cos(\pi/3) = 1/2cos(π/3)=1/2.