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 \(f(x)\). Since \(f(x)=\sin x\), then \(f(2\pi)=\sin(2\pi) = 0\). The limit becomes \(\lim_{x
ightarrow2\pi}\frac{0 - \sin x}{x - 2\pi}=-\lim_{x
ightarrow2\pi}\frac{\sin x}{x - 2\pi}\). Let \(t=x - 2\pi\), then \(x=t + 2\pi\). As \(x
ightarrow2\pi\), \(t
ightarrow0\). So the limit is \(-\lim_{t
ightarrow0}\frac{\sin(t + 2\pi)}{t}\).

Step2: Use trigonometric identity

We know that \(\sin(A + B)=\sin A\cos B+\cos A\sin B\), and \(\sin(t + 2\pi)=\sin t\cos(2\pi)+\cos t\sin(2\pi)=\sin t\) (because \(\cos(2\pi)=1\) and \(\sin(2\pi)=0\)). So the limit is \(-\lim_{t
ightarrow0}\frac{\sin t}{t}\).

Step3: Apply well - known limit

We know the well - known limit \(\lim_{t
ightarrow0}\frac{\sin t}{t}=1\). So \(-\lim_{t
ightarrow0}\frac{\sin t}{t}=- 1\).

Answer:

B. -1