## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to4}\frac{x^2-2x-8}{x-4}=6$$
$$A=\lim_{x\to4}\frac{x^2-2x-8}{x-4}$$ In this exercise, the L'Hospital's Rule is actually not necessary. However, for the sake of everybody, I would carry out both methods, elementary one and L'Hospital's Rule one. 1) Method 1: Elementary method $$A=\lim_{x\to4}\frac{(x^2-4x)+(2x-8)}{x-4}$$ $$A=\lim_{x\to4}\frac{x(x-4)+2(x-4)}{x-4}$$ $$A=\lim_{x\to4}\frac{(x-4)(x+2)}{x-4}$$ $$A=\lim_{x\to4}(x+2)$$ $$A=4+2$$ $$A=6$$ 2) Method 2: L'Hospital's Rule Since $\lim_{x\to4}(x^2-2x-8)=4^2-2\times4-8=16-8-8=0$ and $\lim_{x\to4}(x-4)=4-4=0$, this limit is an indeterminate form of type $\frac{0}{0}$, and we can apply L'Hospital Rule $$A=\lim_{x\to4}\frac{x^2-2x-8}{x-4}$$ $$A=\lim_{x\to4}\frac{\frac{d(x^2-2x-8)}{dx}}{\frac{d(x-4)}{dx}}$$ $$A=\lim_{x\to4}\frac{2x-2}{1}$$ $$A=\lim_{x\to4}(2x-2)$$ $$A=2\times4-2$$ $$A=6$$