Question

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

Answer 1

`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->0) (5+6t^(-1)+t^(-2))/(8-t^(-1)+2t^(-2))* 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`

`=1/2`

Here is the answer :)