Question

My textbook says that `lim x-> - oo (x^2+x+3)/x^2` is >1 from below. I thought is was from above. Which one is correct?

Answer 1

You are correct for being greater than 1 from above. The text is also correct for being greater than 1 from below.

Explanation:

Let's test it.

Let's look at `x=-1`

`((-1)^2-1+3)/(-1)^2=3/1=3`

Let's look at `x=-1/2`

`((-1/2)^2-1/2+3)/(-1/2)^2=(11/4)/(1/4)=11`

The pattern becomes clear - the limit from below is greater than 1.

Keep in mind that you are also right in that the limit from above is also greater than 1 (the squares force both sides to be above 1).

Here's the graph, which might help:

graph{(x^2+x+3)/x^2[-10,10,-1,9]}

Answer 2

The graph approaches the asymptote `y=1` from below.

Explanation:

`lim_(x->-oo) (x^2+x+3)/x^2=lim_(x->-oo) (1+(x+3)/x^2)=1`

The fraction `(x+3)/x^2` approaches 0 as `x->-oo`

In the meanwhile it becomes 0 at `x=-3` and the graph crosses line `y=1` and is below it from now on.

See how it behaves

graph{(y-(x^2+x+3)/x^2)(y-1)=0 [-19, 1, -1, 2]}