Answer
$a)\dfrac {3}{8}$
$b)\dfrac {3}{8}$
Work Step by Step
$a) \lim _{x\rightarrow 4}\dfrac {3\left( x-4\right) }{x^{2}-16}=\dfrac {3\left( x-4\right) }{\left( x-4\right) \left( x+4\right) }=\dfrac {3}{x+4}=\dfrac {3}{4+4}=\dfrac {3}{8}$
$b)\lim _{x\rightarrow 4}\dfrac {3\left( x-4\right) }{x^{2}-16}=\dfrac {\dfrac {d}{dx}\left( 3\left( x-4\right) \right) }{\dfrac {d}{dx}\left( x^{2}-16\right) }=\dfrac {3}{2x}=\dfrac {3}{2\times 4}=\dfrac {3}{8}$
