Answer
$a)\dfrac {3}{2};b)\dfrac {3}{2}$
Work Step by Step
$a)\lim _{x\rightarrow 0}\dfrac {\sin 6x}{4x}=\dfrac {6x}{4x}=\dfrac {6}{4}=\dfrac {3}{2}$
$b)\lim _{x\rightarrow 0}\dfrac {\sin _{6}x}{4x}=\dfrac {\dfrac {d}{dx}\left( \sin _{6}x\right) }{\dfrac {d}{dx}\left( 4x\right) }=\dfrac {6\cos 6x}{4}=\dfrac {6\cos 0}{4}=\dfrac {6}{4}=\dfrac {3}{2}$
