# How do you find lim (5+6t^-1+t^-2)/(8-3t^-1+2t^-2) as t->0?

Dec 5, 2017

$\frac{1}{2}$

#### Explanation:

As the power is negative that it is hard to take the limit, we multiply by ${t}^{2} / {t}^{2}$ to cancel out the negative power.

${\lim}_{t \to 0} \frac{5 + 6 {t}^{- 1} + {t}^{- 2}}{8 - {t}^{- 1} + 2 {t}^{- 2}} \cdot {t}^{2} / {t}^{2}$

=lim_(t->0) (5t^2+6t+1)/(8t^2-t+2

=(5(0)^2+6(0)+1)/(8(0)^2-(0)+2

$= \frac{1}{2}$