Answer
$2\sqrt {6}$
Work Step by Step
$\lim _{x\rightarrow 4}\sqrt {\dfrac {x^{3}-2x^{2}-8x}{x-4}}=\sqrt {\dfrac {x\left( x^{2}-2x-8\right) }{x-4}}=\sqrt {\dfrac {x\left( x-4\right) \left( x+2\right) }{x-4}}=\sqrt {x\left( x+2\right) }=\sqrt {4\times \left( 4+2\right) }=2\sqrt {6}$
