Finding $\lim_{y\to -2}\left(\frac{y^3+8}{y+2}\right)$ Here's my work.
$$\begin{align} \lim_{y\to -2} \;\dfrac{y^3+8}{y+2} &= \lim_{y \to -2}\;\require{cancel}\dfrac{(\cancel{y +2})(y^2 - 2y + 4)}{\cancel{y + 2}}\\ \\
& = \lim_{y \to -2}\;\; y^2 - 2y + 4 \\ \\
& = 4 + 4 + 4 \\ \\
& = 12\end{align}$$
The answer book says that the correct answer is 4
What did I do wrong??

Your answer is entirely correct. There must be a typo in the answer key.
