Question

What is the limit of `[x^2-5x+6]/ [x^3-8]` as x goes to infinity?

Answer 1

`0`

Explanation:

At infinity, the `x^3` term in the denominator dominates the function as it grows much quicker than all the other terms.

Therefore we may write :

`lim_(x->oo)(x^2-5x+6)/(x^3-8)=lim_(x->oo)x^2/x^3=lim_(x->oo)1/x=0`