Solución

lim⁡x→1/22x2+3x−22x−1=12+32−21−1=00\mathop {\lim }\limits_{x \to 1/2} {{2{x^2} + 3x - 2} \over {2x - 1}} = {{{1 \over 2} + {3 \over 2} - 2} \over {1 - 1}} = {0 \over 0}

Así, lim⁡x→1/22x2+3x−22x−1=lim⁡x→1/2(2x−1)(x+2)2x−1=52\mathop {\lim }\limits_{x \to 1/2} {{2{x^2} + 3x - 2} \over {2x - 1}} = \mathop {\lim }\limits_{x \to 1/2} {{\left( {2x - 1} \right)\left( {x + 2} \right)} \over {2x - 1}} = {5 \over 2}