Answer
$\dfrac {-5}{54}$
Work Step by Step
$\lim _{x\rightarrow 3}\dfrac {\sqrt {x+6}-x}{x^{3}-3x^{2}}=\dfrac {\left( \sqrt {x+6}-x\right) \left( \sqrt {x+6}+x\right) }{x^{2}\left( x-3\right) \left( \sqrt {x+6}+x\right) }=\dfrac {x+6-x^{2}}{x^{2}\left( x-3\right) \left( \sqrt {x+6}+x\right) }=\dfrac {-\left( x-3\right) \left( x+2\right) }{x^{2}\left( x-3\right) \left( \sqrt {x+6}+x\right) }=-\dfrac {x+2}{x^{2}\left( \sqrt {x+6}+x\right) }=\dfrac {-5}{54}$
