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<h1>Solución</h1>
$$\mathop {\lim }\limits_{t \to 9} {{t - 9} \over {\sqrt t  - 3}} = {{9 - 9} \over {3 - 3}} = {0 \over 0}$$
<p>Así,
$$\mathop {\lim }\limits_{t \to 9} {{t - 9} \over {\sqrt t  - 3}} = \mathop {\lim }\limits_{t \to 9} {{\left( {t - 9} \right)\left( {\sqrt t  + 3} \right)} \over {\left( {\sqrt t  - 3} \right)\left( {\sqrt t  + 3} \right)}} = \mathop {\lim }\limits_{t \to 9} \left( {\sqrt t  + 3} \right) = 6$$
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