# Asymptotes

## Related calculator: Asymptote Calculator

Definition. The line x=a is called vertical asymptote of the curve y=f(x) if at least one of the following statements is true:

1. lim_(x->a)f(x)=oo
2. lim_(x->a^+)f(x)=oo
3. lim_(x->a^-)f(x)=oo
4. lim_(x->a)f(x)=-oo
5. lim_(x->a^+)f(x)=-oo
6. lim_(x->a^-)f(x)=-oo

For example y-axis (x=0) is a vertical asymptote of the curve y=1/x^2 because lim_(x->0)1/x^2=oo.

Geometrically vertical asymptote is vertical line to which approaches graph of the function but never intersects it. As function approaches a values of function become so large, that we can't draw them. Example 1. Find vertical asymptotes of f(x)=x/(x-1).

In Limits Involving Infinity note we saw that if lim_(x->a^+)f(x)=L!=0 and lim_(x->a+)g(x)=0 then lim_(x->a^+)(f(x)/g(x))=oo (Law 8 for one-sided limits).

So, to find points where function approach infinity, we need to find points where denominator equals 0, and make sure that numerator doesn't equal 0 at this point.

The only point where denominator equals 0 is point 1, and numerator doesn't equal 0 at this point.

Let's find out one-sided limits at this point. When x approaches 1 from the right then x-3 becomes very small positive value and x approaches 1. Thus, lim_(x->1^+)(x)/(x-1)=oo When x approaches 1 from the left then x-1 becomes very small negative value and x approaches 1. Thus, lim_(x->1^-)(x)/(x-1)=-oo. So, we showed that both one-sided limits are infinite. In fact it was sufficiently to show that at least one of the limits is infinite.

Thus, x=1 is vertical asympote.

In fact function can have more than vertical asymptote.

Example 2. Find vertical asymptotes of y=tan(x). Recall that by definition tan(x)=(sin(x))/(cos(x)).

This means that there will be vertical asymptotes at points where cos(x)=0, i.e. points of the form x=pi/2+pik,k in ZZ.

Therefore, tan(x) has infinitely many asymptotes.

Example 3. Find vertical asymptotes of f(x)= (x^2-x)/(x^2+x-2). Since x^2+x-2=(x+2)(x-1) then denominator equals 0 when x=-2 and x=1.

However, at point x=1 numerator also equals 0. We need to additionally check this point.So, lim_(x->1)(x^2-x)/(x^2+x-2)=lim_(x->1)(x(x-1))/((x+2)(x-1))=

=lim_(x->1)x/(x+2)=1/(1+2)=1/3.

Thus, there is only one vertical asymptote x=-2. This example showed, that you should be very cautious: you should check whether numerator equals 0 at points where denominator equals 0.

Definition. Line y=L is called horizontal asymptote of the function f(x) if either lim_(x->oo)f(x)=L or lim_(x->-oo)f(x)=L.

Example 4. Find horizontal asymptotes of y=1+1/x. Since lim_(x->oo)(1+1/x)=1 then line y=1 is horizontal asymptote.

In fact also lim_(x->-oo)(1+1/x)=1, but it suffices one limit to state that y=1 is horizontal asymptote.

In general function can have more than one horizontal asymptote.

Example 5. Find horizontal asymptotes of y=arctan(x). Since lim_(x->oo)arctan(x)=pi/2 and lim_(x->-oo)arctan(x)=-pi/2 then there are two vertical asymptotes: y=pi/2 and y=-pi/2.

Definition. Line y=mx+b is called slant (oblique) asymptote of function y=f(x) if either lim_(x->oo)(f(x)-(mx+b))=0 or lim_(x->-oo)(f(x)-(mx+b))=0.

We calculate m as follows: m=lim_(x->oo)(f(x))/x (m=lim_(x->-oo)(f(x))/x).

If this limit is not finite or doesn't exist then there is no oblique asymptote.

b is calculated as follows: b=lim_(x->oo)(f(x)-mx)(b=lim_(x->-oo)(f(x)-mx)).

Example 6. Find all asymptotes of y=x+1/x. There are no horizontal asymptotes because lim_(x->oo)(x+1/x)=oo and lim_(x->-oo)(x+1/x)=-oo.

There is vertical asymptote x=0 because lim_(x->0^+)(x+1/x)=oo.

This function also has oblique asymptote. Indeed, m=lim_(x->oo)(x+1/x)/x=lim_(x->oo)(1+1/x^2)=1.

So, m=1. Now, b=lim_(x->oo)(x+1/x-1*x)=lim_(x->oo)1/x=0. So, b=0.

Therefore, slant asymptote is y=x. On the figure to the right blue line is vertical asymptote x=0, green line is slant asymptote y=x.

Note, that we will obtain same slant asymptote when x->-oo.

In general, it is a good practice to treat cases x->oo and x->-oo separately.