# How do you find lim (3t^3+4)/(t^2+t-2) as t->1?

${\lim}_{t \to 1} \frac{3 {t}^{3} + 4}{{t}^{2} + t - 2} = \infty$
You simply could substitute $t = 1$ in $\frac{3 {t}^{3} + 4}{{t}^{2} + t - 2}$ and get:
$\frac{3 \cdot {1}^{3} + 4}{{1}^{2} + 1 - 2} = \frac{7}{0} = \infty$