Answer
$-\dfrac{1}{3}$
Work Step by Step
Given: $\lim\limits_{x\to1}\dfrac{x^3-2x^2+1}{x^3-1}$
This shows an indeterminate form of type $\dfrac{0}{0}$, and so we will apply L'Hospital Rule.
$\lim\limits_{x\to1}\frac{\dfrac{d(x^3-2x^2+1)}{dx}}{\dfrac{d(x^3-1)}{dx}}=\lim\limits_{x\to1}\dfrac{3x^2-4x}{3x^2}$
or, $=\dfrac{3\times1^2-4\times1}{3\times1^2}$
or, $=\dfrac{3-4}{3}$
or,$ =-\dfrac{1}{3}$
