Solución

lim⁡x→4x2−16x−4=16−164−4=00\mathop {\lim }\limits_{x \to 4} {{{x^2} - 16} \over {x - 4}} = {{16 - 16} \over {4 - 4}} = {0 \over 0}

Así, lim⁡x→4x2−16x−4=lim⁡t→4(x+4)(x−4)x−4=8\mathop {\lim }\limits_{x \to 4} {{{x^2} - 16} \over {x - 4}} = \mathop {\lim }\limits_{t \to 4} {{\left( {x + 4} \right)\left( {x - 4} \right)} \over {x - 4}} = 8