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<h1>Solución</h1>
$$\mathop {\lim }\limits_{x \to 4} {{{x^2} - 16} \over {x - 4}} = {{16 - 16} \over {4 - 4}} = {0 \over 0}$$
<p>Así,
$$\mathop {\lim }\limits_{x \to 4} {{{x^2} - 16} \over {x - 4}} = \mathop {\lim }\limits_{t \to 4} {{\left( {x + 4} \right)\left( {x - 4} \right)} \over {x - 4}} = 8$$
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