# How do you find lim (1-5t^-1)/(4+6t^-1) as t->0?

$\frac{5}{6}$
Since ${t}^{-} 1 = \frac{1}{t}$ you could rewrite:
${\lim}_{t \to 0} \frac{1 - \frac{5}{t}}{4 - \frac{6}{t}} = {\lim}_{t \to 0} \frac{\frac{t - 5}{\cancel{t}}}{\frac{4 t - 6}{\cancel{t}}} = {\lim}_{t \to 0} \frac{t - 5}{4 t - 6} = \frac{5}{6}$