Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to-2}\frac{x^3+8}{x+2}=12$$
$$A=\lim_{x\to-2}\frac{x^3+8}{x+2}$$ I will still carry out both methods: elementary one and L'Hospital's Rule one, since elementary method is still applicable here. 1) Method 1: Elementary method We expand $(x^3+8)$ using the sum of cubes formula: $(a^3+b^3)=(a+b)(a^2-ab+b^2)$. $$A=\lim_{x\to-2}\frac{(x+2)(x^2-2x+4)}{x+2}$$ $$A=\lim_{x\to-2}(x^2-2x+4)$$ $$A=(-2)^2-2(-2)+4$$ $$A=4+4+4=12$$ 2) Method 2: L'Hospital's Rule Since $\lim_{x\to-2}(x^3+8)=(-2)^3+8=-8+8=0$ and $\lim_{x\to-2}(x+2)=-2+2=0$, this limit is an indeterminate form of type $\frac{0}{0}$, and we can apply L'Hospital Rule $$A=\lim_{x\to-2}\frac{\frac{d(x^3+8)}{dx}}{\frac{d(x+2)}{dx}}$$ $$A=\lim_{x\to-2}\frac{3x^2}{1}$$ $$A=\lim_{x\to-2}(3x^2)$$ $$A=3\times(-2)^2$$ $$A=3\times4=12$$