if (f(x)=sin x), then (lim_{x\rightarrow2pi}\frac{f(2pi)-f(x)}{x - 2pi}=) \na (-2pi) \nb (-1) \nc (1) \nd…

Explanation:

Step1: Substitute function values

First, find (f(2\pi)) and (f(x)). Given (f(x)=\sin x), then (f(2\pi)=\sin(2\pi) = 0). So the limit becomes (\lim_{x\rightarrow2\pi}\frac{0 - \sin x}{x - 2\pi}=-\lim_{x\rightarrow2\pi}\frac{\sin x}{x - 2\pi}).

Step2: Use limit - derivative relationship

Recall the definition of the derivative (f^\prime(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x - a}). Let (a = 2\pi) and (f(x)=\sin x), and we know that the derivative of (y = \sin x) is (y^\prime=\cos x). So (-\lim_{x\rightarrow2\pi}\frac{\sin x}{x - 2\pi}=-\cos(2\pi)).

Step3: Evaluate the cosine value

Since (\cos(2\pi)=1), then (-\cos(2\pi)=- 1).

Answer:

B. -1