Answer
$+\infty$
Work Step by Step
When $x$ approaches $3$ from the left, $3-x$ is positive, so we can write:
$3-x=\sqrt{(3-x)^{2}}$
Thus:
$\displaystyle \lim_{x\rightarrow 3^{-}}\frac{\sqrt{3-x}}{3-x}=\lim_{x\rightarrow 3^{-}}\frac{\sqrt{3-x}}{\sqrt{(3-x)^{2}}}=\lim_{x\rightarrow 3^{-}}\sqrt{\frac{3-x}{(3-x)^{2}}}=\sqrt{\lim_{x\rightarrow 3^{-}}\frac{1}{3-x}}$
The form of $\displaystyle \lim_{x\rightarrow 3^{-}}\frac{1}{3-x}$ is $\displaystyle \frac{k}{0^{+}}=\pm\infty \qquad \frac{k}{Small}=Big $
and $\sqrt{Big}=Big\qquad$
Thus, we have:
$\displaystyle \lim_{x\rightarrow 3^{-}}\frac{\sqrt{3-x}}{3-x}=+\infty$
