Answer
$$\lim _{y \rightarrow 3} \frac{y^{2}+y-12}{y^{3}-10 y+3} =\frac{7}{17}$$
Work Step by Step
Given $$\lim _{y \rightarrow 3} \frac{y^{2}+y-12}{y^{3}-10 y+3}$$
let $$ f(y) = \frac{y^{2}+y-12}{y^{3}-10 y+3}$$
Since, we have
$$ f(3)= \frac{9+3-12}{27-30+3}=\frac{0}{0}$$
So, transform algebraically and cancel
\begin{aligned}L&=\lim _{y \rightarrow 3} \frac{y^{2}+y-12}{y^{3}-10 y+3}\\
&=\lim _{y \rightarrow 3} \frac{(y-3)(y+4)}{(y-3)\left(y^{2}+3 y-1\right)}\\
&=\lim _{y \rightarrow 3} \frac{(y+4)}{\left(y^{2}+3 y-1\right)}\\
&=\frac{3+4}{9+9-1}\\
&=\frac{7}{17}
\end{aligned}
