Question

Question #4aef3

Answer 1

Read below.

Explanation:

So we have `lim_(x->oo)(x^2-18x)/(x+5)`

Let's use our logic here.
As `x` gets really, really, large, only the largest term in both the numerator and the denominator will matter.

For example, in `x^2-18x`, `x^2` will be so large that `18x` will be as good as 0.

We can therefore say that when `x` is `oo` in `(x^2-18x)/(x+5)`, the answer will be very similar as `x^2/x`.

We simplify this to get: `x`.

Since we are saying that `x` is infinity, this means that as `x` approaches `oo`, the `y` will get unboundedly large, or `oo`