Answer
$\lim\limits_{x \to 1}\frac{3x+2}{2x+3}$ = 1
Work Step by Step
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
