Question

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

Answer 1

`lim_(t->1)(3t^3+4)/(t^2+t-2)=oo`

Explanation:

You simply could substitute `t=1` in `(3t^3+4)/(t^2+t-2)` and get:

`(3*1^3+4)/(1^2+1-2)=7/0=oo`