Question

When f(x) is a differentiable function

satisfying
`\lim_(x \to \\\pi ) (f(x))/(x-\pi)`
Find
`\lim_(x \to \\\pi ) (sin(2f(x)))/(x-\pi)`

Answer 1

Presumably, the first limit is some finite number

`\displaystyle \lim_(x\to\pi) (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) (\sin(2 f(x)))/(x - \pi) = \lim_(x\to\pi) (2 \sin(f(x)) \cos(f(x)))/(x - \pi) = 2 \lim_(x\to\pi) (\sin(f(x)))/(f(x)) * (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)(\sin(f(x)))/(f(x)) = \lim_(f(x)\to0)(\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) (\sin(f(x)))/(f(x)) * (f(x) \cos(f(x)))/(x - \pi) \\\\ = 2 \left(\lim_(x\to\pi)(\sin(f(x)))/(f(x))\right) \left(\lim_(x\to\pi) (f(x))/(x-\pi)\right) \left(\lim_(x\to\pi)\cos(f(x))\right) \\\\ = 2 * 1 * c * 1 = \boxed{2c}`