Answer
$\lim\limits_{x \to 3^-}\frac{\sqrt x}{(x-3)^5}=-\infty$
Work Step by Step
$\lim\limits_{x \to 3^-}\frac{\sqrt x}{(x-3)^5}=\frac{\sqrt3^-}{(0^-)^5}=\frac{\sqrt3^-}{0^-}=-\infty$
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