Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to1}\frac{x^3-2x^2+1}{x^3-1}=-\frac{1}{3}$$
$$A=\lim_{x\to1}\frac{x^3-2x^2+1}{x^3-1}$$ 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 - For the numerator: $$x^3-2x^2+1$$ $$=(x^3-x^2)+(-x^2+x)+(-x+1)$$ $$=x^2(x-1)-x(x-1)-(x-1)$$ $$=(x-1)(x^2-x-1)$$ - For the denominator: $$x^3-1=(x-1)(x^2+x+1)$$ Therefore, $$A=\lim_{x\to1}\frac{(x-1)(x^2-x-1)}{(x-1)(x^2+x+1)}$$ $$A=\lim_{x\to1}{\frac{x^2-x-1}{x^2+x+1}}$$ $$A=\frac{1^2-1-1}{1^2+1+1}$$ $$A=\frac{-1}{3}$$ 2) Method 2: L'Hospital's Rule Since $\lim_{x\to1}(x^3-2x^2+1)=1^3-2\times1^2+1=1-2+1=0$ and $\lim_{x\to1}(x^3-1)=1^3-1=0$, this limit is an indeterminate form of type $\frac{0}{0}$, and we can apply L'Hospital Rule $$A=\lim_{x\to1}\frac{\frac{d(x^3-2x^2+1)}{dx}}{\frac{d(x^3-1)}{dx}}$$ $$A=\lim_{x\to1}\frac{3x^2-4x}{3x^2}$$ $$A=\frac{3\times1^2-4\times1}{3\times1^2}$$ $$A=\frac{3-4}{3}$$ $$A=-\frac{1}{3}$$