\(\lim\limits_{x\rightarrow0}\left(\dfrac{1}{x}+\dfrac{1}{x^2}\right)=\infty+\infty=\infty\)
\(\lim\limits_{x\rightarrow-2}\left(\dfrac{x^3+8}{x+2}\right)=\lim\limits_{x\rightarrow-2}\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x+2}=\lim\limits_{x\rightarrow-2}\left(x^2-2x+4\right)=12\)