Question

How do you find the horizontal asymptote of a curve?

Answer 1

To find the horizontal asymptote (generally of a rational function), you will need to use the Limit Laws, the definitions of limits at infinity, and the following theorem: `lim_(x->oo) (1/x^r) = 0` if `r` is rational, and `lim_(x->-oo) (1/x^r) = 0` if `r` is rational and `x^r` is defined.

Recall from the definition of limits that we can only take limits of real numbers and infinity is not a real number, which is why we need the previous theorem.

The strategy for using the theorem is to divide every term by the highest power term from the denominator; this should leave us with a polynomial in the numerator or a constant. If we have a polynomial, then there is no horizontal asymptote. If we have a constant, then y=constant is our horizontal asymptote.

For example:

`lim_(x->oo) (3x^2-4x+8)/(7x^2+5x-9)`

We divded every term by `x^2`.

`=lim_(x->oo) (3x^2/x^2-4x/x^2+8/x^2)/(7x^2/x^2+5x/x^2-9/x^2)`

Now use our limit laws.

`=(lim_(x->oo) 3x^2/x^2-lim_(x->oo) 4x/x^2+lim_(x->oo) 8/x^2)/(lim_(x->oo) 7x^2/x^2+lim_(x->oo) 5x/x^2-lim_(x->oo) 9/x^2)`

Now simplify.

`=(lim_(x->oo) 3-lim_(x->oo) 4/x+lim_(x->oo) 8/x^2)/(lim_(x->oo) 7+lim_(x->oo) 5/x-lim_(x->oo) 9/x^2)`

Finally use the theorem and the limit law of a constant.

`=(3-0+0)/(7+0-0)`

`=3/7`

So, in this case we have a horizontal asymptote of `y=3/7`.

If we ended up with `-4x^2+11x-12` in the numerator, then there would be no horizontal asymptote as the function grow unbounded negatively.

The theorem @ `-oo` has an extra condition because `x^r` must be defined, that is if `r` is rational, then the denominator must be odd. In case you forgot, `sqrt(-2)` is not defined.

This is the general strategy, obviously more difficult questions can be framed and you should read your textbook for those examples.