Question

When f(x) is a differentiable function

Answer 1

Presumably, the first limit is some finite number

$\displaystyle \lim_{x\to\pi} \frac{f(x)}{x-\pi} = c$

Since x - π clearly approaches 0 as x approaches π, we must also have f(x) approaching 0,

$\displaystyle \lim_{x\to\pi} f(x) = 0$

Recall the double angle identity,

sin(2t) = 2 sin(t) cos(t)

and rewrite the limit as

$\displaystyle \lim_{x\to\pi} \frac{\sin(2 f(x))}{x - \pi} = \lim_{x\to\pi} \frac{2 \sin(f(x)) \cos(f(x))}{x - \pi} = 2 \lim_{x\to\pi} \frac{\sin(f(x))}{f(x)} \times \frac{f(x) \cos(f(x))}{x - \pi}$

Recall that

$\displaystyle \lim_{x\to0}\frac{\sin(x)}x=1$

which means

$\displaystyle \lim_{x\to\pi}\frac{\sin(f(x))}{f(x)} = \lim_{f(x)\to0}\frac{\sin(f(x))}{f(x)} = 1$

and by continuity,

$\displaystyle \lim_{x\to\pi} \cos(f(x)) = \cos\left(\lim_{x\to\pi} f(x)\right) = \cos(0) = 1$

Then

$\displaystyle 2 \lim_{x\to\pi} \frac{\sin(f(x))}{f(x)} \times \frac{f(x) \cos(f(x))}{x - \pi} \\\\ = 2 \left(\lim_{x\to\pi}\frac{\sin(f(x))}{f(x)}\right) \left(\lim_{x\to\pi} \frac{f(x)}{x-\pi}\right) \left(\lim_{x\to\pi}\cos(f(x))\right) \\\\ = 2 \times 1 \times c \times 1 = \boxed{2c}$