Question

How do you evaluate `(t^3-2t+3)/ ( 5-2t^2 )` as t approaches infinity?

Answer 1

`-oo`

Explanation:

`lim_(x->oo) (t^3-2t+3)/(5-2t^2)`

`=lim_(x->oo) t^3/(-2t^2)`-> pick the highest degree term from the top and the bottom

`=lim_(x->oo) -t/2`

`=-oo`