## Calculus, 10th Edition (Anton)

$\lim\limits_{x \to 1}\frac{3x+2}{2x+3}$ = 1
3x² - x - 2 = (x - 1)(3x + 2) 2x² + x - 3 = (x - 1)(2x + 3), so: $\frac{3x² - x - 2}{2x² + x - 3}$ = $\frac{(x-1)(3x+2)}{(x-1)(2x+3)}$ = $\frac{3x+2}{2x+3}$ $\lim\limits_{x \to 1}\frac{3x² - x - 2}{2x² + x - 3}$ = $\lim\limits_{x \to 1}\frac{3x+2}{2x+3}$ = $\frac{3\times1+2}{2\times1+3}$ = 1