Question

How do you find `lim (1-5t^-1)/(4+6t^-1)` as `t->0`?

Answer 1

`5/6`

Explanation:

Since `t^-1=1/t` you could rewrite:

`lim_(t->0) (1-5/t)/(4-6/t)=lim_(t->0) ((t-5)/cancelt)/((4t-6)/cancelt)=lim_(t->0)(t-5)/(4t-6)=5/6`