Así, $$\mathop {\lim }\limits_{t \to 9} {{t - 9} \over {\sqrt t - 3}} = \mathop {\lim }\limits_{t \to 9} {{\left( {t - 9} \right)\left( {\sqrt t + 3} \right)} \over {\left( {\sqrt t - 3} \right)\left( {\sqrt t + 3} \right)}} = \mathop {\lim }\limits_{t \to 9} \left( {\sqrt t + 3} \right) = 6$$