Question

if (f(x)=sin x), then (lim_{x
ightarrow2pi}\frac{f(2pi)-f(x)}{x - 2pi}=)
a (-2pi)
b (-1)
c (1)
d (2pi)

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 \(\lim_{x
ightarrow2\pi}\frac{0 - \sin x}{x - 2\pi}=-\lim_{x
ightarrow2\pi}\frac{\sin x}{x - 2\pi}\).

Step2: Use limit - derivative relationship

Recall the definition of the derivative \(f^\prime(a)=\lim_{x
ightarrow a}\frac{f(x)-f(a)}{x - a}\). 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 \(-\lim_{x
ightarrow2\pi}\frac{\sin x}{x - 2\pi}=-\cos(2\pi)\).

Step3: Evaluate the cosine value

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

Answer:

B. -1