Find ${\lim }_{x\to 1}\frac{x^2-1}{x^2+x-2}$.

Solution

1. Factor the numerator: $x^2-1=(x-1)(x+1)$.
2. Factor the denominator: $x^2+x-2=(x-1)(x+2)$.
3. Simplify the fraction: $\frac{(x-1)(x+1)}{(x-1)(x+2)}=\frac{x+1}{x+2}$ for $x\ne 1$.
4. Evaluate the limit: $$\begin{array}{rl}\underset{x\to 1}{\lim }\frac{x+1}{x+2}& =\frac{1+1}{1+2}\\ & =\frac{2}{3}\end{array}$$.
5. Hence the answer is (B).