Answer
$$\lim_{x\to1}\frac{x^3-2x^2+1}{x^3-1}=-\frac{1}{3}$$
Work Step by Step
$$A=\lim_{x\to1}\frac{x^3-2x^2+1}{x^3-1}$$
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
$$A=\lim_{x\to1}\frac{\frac{d(x^3-2x^2+1)}{dx}}{\frac{d(x^3-1)}{dx}}$$
$$A=\lim_{x\to1}\frac{3x^2-4x}{3x^2}$$
$$A=\frac{3\times1^2-4\times1}{3\times1^2}$$
$$A=\frac{3-4}{3}$$
$$A=-\frac{1}{3}$$
