Statement Calculate limx→1x3−1x2−1\displaystyle\lim_{x\to 1}\frac{x^3-1}{x^2-1}x→1limx2−1x3−1. Solution Substituting x=1x=1x=1 gives 0/00/00/0. We factorise using differences of powers: x3−1=(x−1)(x2+x+1),x2−1=(x−1)(x+1).x^3-1 = (x-1)(x^2+x+1), \qquad x^2-1 = (x-1)(x+1).x3−1=(x−1)(x2+x+1),x2−1=(x−1)(x+1). We cancel the common factor (x−1)(x-1)(x−1): x3−1x2−1=(x−1)(x2+x+1)(x−1)(x+1)=x2+x+1x+1.\frac{x^3-1}{x^2-1} = \frac{(x-1)(x^2+x+1)}{(x-1)(x+1)} = \frac{x^2+x+1}{x+1}.x2−1x3−1=(x−1)(x+1)(x−1)(x2+x+1)=x+1x2+x+1. Now we substitute x=1x=1x=1: limx→1x2+x+1x+1=1+1+11+1=32.\lim_{x\to 1}\frac{x^2+x+1}{x+1} = \frac{1+1+1}{1+1} = \boxed{\frac{3}{2}}.limx→1x+1x2+x+1=1+11+1+1=23. Links Topics: Limits Concepts: Indeterminate forms Skills: Calculating limits · Factorising Exercise type: Limit calculation