Solución

lim⁡x→63x−182x−12=18−1812−12=00\mathop {\lim }\limits_{x \to 6} {{3x - 18} \over {2x - 12}} = {{18 - 18} \over {12 - 12}} = {0 \over 0}

Así, lim⁡x→63x−182x−12=lim⁡t→93(x−6)2(x−6)=32\mathop {\lim }\limits_{x \to 6} {{3x - 18} \over {2x - 12}} = \mathop {\lim }\limits_{t \to 9} {{3\left( {x - 6} \right)} \over {2\left( {x - 6} \right)}} = {3 \over 2}