# How do you find the horizontal asymptote of a curve?

##### 1 Answer
Aug 14, 2014

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: ${\lim}_{x \to \infty} \left(\frac{1}{x} ^ r\right) = 0$ if $r$ is rational, and ${\lim}_{x \to - \infty} \left(\frac{1}{x} ^ r\right) = 0$ 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:

${\lim}_{x \to \infty} \frac{3 {x}^{2} - 4 x + 8}{7 {x}^{2} + 5 x - 9}$

We divded every term by ${x}^{2}$.

$= {\lim}_{x \to \infty} \frac{3 {x}^{2} / {x}^{2} - 4 \frac{x}{x} ^ 2 + \frac{8}{x} ^ 2}{7 {x}^{2} / {x}^{2} + 5 \frac{x}{x} ^ 2 - \frac{9}{x} ^ 2}$

Now use our limit laws.

$= \frac{{\lim}_{x \to \infty} 3 {x}^{2} / {x}^{2} - {\lim}_{x \to \infty} 4 \frac{x}{x} ^ 2 + {\lim}_{x \to \infty} \frac{8}{x} ^ 2}{{\lim}_{x \to \infty} 7 {x}^{2} / {x}^{2} + {\lim}_{x \to \infty} 5 \frac{x}{x} ^ 2 - {\lim}_{x \to \infty} \frac{9}{x} ^ 2}$

Now simplify.

$= \frac{{\lim}_{x \to \infty} 3 - {\lim}_{x \to \infty} \frac{4}{x} + {\lim}_{x \to \infty} \frac{8}{x} ^ 2}{{\lim}_{x \to \infty} 7 + {\lim}_{x \to \infty} \frac{5}{x} - {\lim}_{x \to \infty} \frac{9}{x} ^ 2}$

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

$= \frac{3 - 0 + 0}{7 + 0 - 0}$

$= \frac{3}{7}$

So, in this case we have a horizontal asymptote of $y = \frac{3}{7}$.

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

The theorem @ $- \infty$ 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.