Limits at Infinity and Horizontal Asymptotes

At the beginning of this section we briefly considered what happens to (f(x) = 1/x^2) as (x) grew very large. Graphically, it concerns the behavior of the function to the “far right” of the graph. We make this notion more explicit in the following definition.

We can also define limits such as (limlimits_{xrightarrowinfty}f(x)=infty) by combining this definition with Definition 5.

Horizontal asymptotes can take on a variety of forms. Figure 1.36(a) shows that (f(x) = x/(x^2+1)) has a horizontal asymptote of (y=0), where 0 is approached from both above and below.

Figure 1.36(b) shows that (f(x) =x/sqrt{x^2+1}) has two horizontal asymptotes; one at (y=1) and the other at (y=-1).

Figure 1.36(c) shows that (f(x) = (sin x)/x) has even more interesting behavior than at just (x=0); as (x) approaches (pminfty), (f(x)) approaches 0, but oscillates as it does this.

(text{FIGURE 1.36}): Considering different types of horizontal asymptotes.

We can analytically evaluate limits at infinity for rational functions once we understand (limlimits_{xrightarrowinfty} 1/x). As (x) gets larger and larger, the (1/x) gets smaller and smaller, approaching 0. We can, in fact, make (1/x) as small as we want by choosing a large enough value of (x). Given (epsilon), we can make (1/x<epsilon) by choosing (x>1/epsilon). Thus we have (limlimits_{xrightarrowinfty} 1/x=0).

It is now not much of a jump to conclude the following:

[limlimits_{xrightarrowinfty}frac1{x^n}=0quad text{and}quad limlimits_{xrightarrow-infty}frac1{x^n}=0]

Now suppose we need to compute the following limit:

[limlimits_{xrightarrowinfty}frac{x^3+2x+1}{4x^3-2x^2+9}.]

A good way of approaching this is to divide through the numerator and denominator by (x^3) (hence dividing by 1), which is the largest power of (x) to appear in the function. Doing this, we get

[begin{align*}limlimits_{xrightarrowinfty}frac{x^3+2x+1}{4x^3-2x^2+9} &=limlimits_{xrightarrowinfty}frac{1/x^3}{1/x^3}cdotfrac{x^3+2x+1}{4x^3-2x^2+9} &=limlimits_{xrightarrowinfty}frac{x^3/x^3+2x/x^3+1/x^3}{4x^3/x^3-2x^2/x^3+9/x^3} &= limlimits_{xrightarrowinfty}frac{1+2/x^2+1/x^3}{4-2/x+9/x^3}.end{align*}]

Then using the rules for limits (which also hold for limits at infinity), as well as the fact about limits of (1/x^n), we see that the limit becomes[frac{1+0+0}{4-0+0}=frac14.]

This procedure works for any rational function. In fact, it gives us the following theorem.

We can see why this is true. If the highest power of (x) is the same in both the numerator and denominator (i.e. (n=m)), we will be in a situation like the example above, where we will divide by (x^n) and in the limit all the terms will approach 0 except for (a_nx^n/x^n) and (b_mx^m/x^n). Since (n=m), this will leave us with the limit (a_n/b_m). If (n<m), then after dividing through by (x^m), all the terms in the numerator will approach 0 in the limit, leaving us with (0/b_m) or 0. If (n>m), and we try dividing through by (x^n), we end up with all the terms in the denominator tending toward 0, while the (x^n) term in the numerator does not approach 0. This is indicative of some sort of infinite limit.

Intuitively, as (x) gets very large, all the terms in the numerator are small in comparison to (a_nx^n), and likewise all the terms in the denominator are small compared to (b_nx^m). If (n=m), looking only at these two important terms, we have ((a_nx^n)/(b_nx^m)). This reduces to (a_n/b_m). If (n<m), the function behaves like (a_n/(b_mx^{m-n})), which tends toward 0. If (n>m), the function behaves like (a_nx^{n-m}/b_m), which will tend to either (infty) or (-infty) depending on the values of (n), (m), (a_n), (b_m) and whether you are looking for (limlimits_{xrightarrowinfty} f(x)) or (limlimits_{xrightarrow-infty} f(x)).

With care, we can quickly evaluate limits at infinity for a large number of functions by considering the largest powers of (x). For instance, consider again (limlimits_{xtopminfty}frac{x}{sqrt{x^2+1}},) graphed in Figure ref{fig:hzasy}(b). When (x) is very large, (x^2+1 approx x^2). Thus [sqrt{x^2+1}approx sqrt{x^2} = |x|,quad text{and}quad frac{x}{sqrt{x^2+1}} approx frac{x}{|x|}.]This expression is 1 when (x) is positive and (-1) when (x) is negative. Hence we get asymptotes of (y=1) and (y=-1), respectively.