Answer
$\infty $
Work Step by Step
$\lim _{x\rightarrow 3^-}\dfrac {x-4}{x^{2}-3x}=\lim _{x\rightarrow 3}\dfrac {x-4}{x\left( x-3\right) }=\dfrac {3^--4}{3\left( 3^--3\right) }=\dfrac {-1}{3\times 0^-}=\dfrac {-1}{0^-}=\infty $
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