Question

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

Answer 1

`0`

Explanation:

Divide terms on numerator/denominator by the highest power of t, that is `t^2`

`rArrlim_(xtooo)(((5t)/t^2+2/t^2)/(t^2/t^2-(6t)/t^2+1/t^2))`

`=lim_(xtooo)((5/t+2/t^2)/(1-6/t+1/t^2))`

`=(0+0)/(1-0+0)=0`