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Question: \[ f(x) = \begin{cases} -1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{cases} \] is...

f(x)={−1,x<00,x=01,x>0f(x) = \begin{cases} -1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{cases}

is

A

f(g(x))f(g(x))

B

f(x)f(x)

C

g(x)g(x)

D

f(x)=2x+13x−2,f(x) = \frac{2x + 1}{3x - 2},

Answer

f(x)f(x)

Explanation

Solution

lim⁡x→π32sin⁡(x−π3)1−2cos⁡x\lim _ { x \rightarrow \frac { \pi } { 3 } } \frac { 2 \sin \left( x - \frac { \pi } { 3 } \right) } { 1 - 2 \cos x } = lim⁡x→π32cos⁡(x−π3)2sin⁡x\lim _ { x \rightarrow \frac { \pi } { 3 } } \frac { 2 \cos \left( x - \frac { \pi } { 3 } \right) } { 2 \sin x }

[ L' Hopital Rule]

= cos⁡0/sin⁡(π/3)=23\cos 0 / \sin ( \pi / 3 ) = \frac { 2 } { \sqrt { 3 } }