Answer
$$\lim _{x \rightarrow-2}\left(3 x^{2}+e^{x}\right) =\frac{12e^2+1}{e^2}$$
Work Step by Step
Given: $$\lim _{x \rightarrow-2}\left(3 x^{2}+e^{x}\right)$$ By using the method of replacement: \begin{align*} \lim _{x \rightarrow-2}\left(3 x^{2}+e^{x}\right)&=\lim _{x \rightarrow-2}\left(3 (-2)^{2}+e^{-2}\right)\\
&=\lim _{x \rightarrow-2}\left(12+e^{-2}\right)\\
&=\frac{12e^2+1}{e^2}
\end{align*}
