Question

Find the following limit.
$\underset{x\to 1}{lim}\frac{x^3+2x-3}{x^2+x-2}.$

Answer 1

$\underset{x\to 1}{lim}\frac{x^3+2x-3}{x^2+x-2}$

$=\underset{x\to 1}{lim}\frac{(x-1)(x^2+x+3)}{x^2+2x-x-2}$

$=\underset{x\to 1}{lim}\frac{(x-1)(x^2+x+3)}{x(x+2)-1(x+2)}$

$=\underset{x\to 1}{lim}\frac{(x-1)(x^2+x+3)}{(x-1)(x+2)}$

$=\frac{1+1+3}{2+1}=\frac{5}{2+1}=\frac{5}{3}$.