Answer
$\dfrac {3}{10}$
Work Step by Step
$\lim _{x\rightarrow 3}\dfrac {\sqrt {3x+16}-5}{x-3}=\lim _{x\rightarrow 3}\dfrac {\left( \sqrt {3x+16}-5\right) \left( \sqrt {3x+16}+5\right) }{\left( x-3\right) \left( \sqrt {3x+16}+5\right) }=\lim _{x\rightarrow 3}\dfrac {\left( \sqrt {3x+16}\right) ^{2}-5^2}{\left( x-3\right) \left( \sqrt {x+16}+5\right) }=\lim _{x\rightarrow 3}\dfrac {3x+16-25}{\left( x-3\right) \left( \sqrt {3x+16}+5\right) }=\lim _{x\rightarrow 3}\dfrac {3\left( x-3\right) }{\left( x-3\right) \left( \sqrt {3x+16}+5\right) }=\lim _{x\rightarrow 3}\dfrac {3}{\sqrt {3x+16}+5}=\dfrac {3}{\sqrt {3\times 3+16}+5}=\dfrac {3}{5+5}=\dfrac {3}{10}$
