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->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.