Answer
$a)\dfrac {1}{8};b)\dfrac {1}{8}$
Work Step by Step
$a)\lim _{x\rightarrow 6}\dfrac {\sqrt {x+10}-4}{x-6}=\dfrac {\left( \sqrt {x+10}-4\right) \left( \sqrt {x+10}+4\right) }{\left( x-6\right) \left( \sqrt {x+10}+4\right) }=\dfrac {x-6}{\left( x-6\right) \left( \sqrt {x+10}+4\right) }=\dfrac {1}{\sqrt {x+10}+4}=\dfrac {1}{8}$
$b)\lim _{x\rightarrow 6}\dfrac {\sqrt {x+10}-4}{x-6}=\dfrac {\dfrac {d}{dx}\left( \sqrt {x+10}-4\right) }{\dfrac {d}{dx}\left( x-6\right) }=\dfrac {1}{\dfrac {2\sqrt {x+10}}{1}}=\dfrac {1}{2\sqrt {6+10}}=\dfrac {1}{8}$
