Answer
$$ \frac{5}{6}.$$
Work Step by Step
We have
$$
\lim _{x \rightarrow 1} \frac{\sqrt{8+x}-3 x^{1 / 3}}{x^{2}-3 x+2}= \frac{3-3 }{1-3 +2}=\frac{0}{0}.
$$
is an intermediate form, then we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 1} \frac{\sqrt{8+x}-3 x^{1 / 3}}{x^{2}-3 x+2}= \lim _{x \rightarrow 1} \frac{(1/(2\sqrt{8+x}))- x^{-2 / 3}}{2x-3 }=\frac{(1/6) -1 }{2-3 }=\frac{-5/6}{-1}=\frac{5}{6}.
$$
