If the limit of \(f(x)\) as \(x\) approaches \(c\) from either the left or right (or both) is \(\infty\) or \(-\infty\text{,}\) we say the function has a vertical asymptote at \(c\text{.}\)

When a rational function has a vertical asymptote at \(x=c\text{,}\) we can conclude that the denominator is \(0\) at \(x=c\text{.}\) However, just because the denominator is \(0\) at a certain point does not mean there is a vertical asymptote there. For instance, \(f(x)=(x^2-1)/(x-1)\) does not have a vertical asymptote at \(x=1\text{,}\) as shown in Figure 1.6.9. While the denominator does get small near \(x=1\text{,}\) the numerator gets small too, matching the denominator step for step. In fact, factoring the numerator, we get \begin{equation*} f(x)=\frac{(x-1)(x+1)}{x-1}. \end{equation*}

Canceling the common term, we get that \(f(x)=x+1\) for \(x\not=1\text{.}\) So there is clearly no asymptote, rather a hole exists in the graph at \(x=1\text{.}\)

The above example may seem a little contrived. Another example demonstrating this important concept is \(f(x)= (\sin(x) )/x\text{.}\) We have considered this function several times in the previous sections. We found that \(\lim\limits_{x\to0}\frac{\sin(x) }{x}=1\text{;}\) i.e., there is no vertical asymptote. No simple algebraic cancellation makes this fact obvious; we used the 1.3.8 in Section 1.3 to prove this.

If the denominator is \(0\) at a certain point but the numerator is not, then there will usually be a vertical asymptote at that point. On the other hand, if the numerator and denominator are both zero at that point, then there may or may not be a vertical asymptote at that point. This case where the numerator and denominator are both zero returns us to an important topic.

We have seen how the limits \(\lim\limits_{x\to 0}\frac{\sin(x) }{x}\) and \(\lim\limits_{x\to1}\frac{x^2-1}{x-1}\) each return the indeterminate form \(0/0\) when we blindly plug in \(x=0\) and \(x=1\text{,}\) respectively. However, \(0/0\) is not a valid arithmetical expression. It gives no indication that the respective limits are \(1\) and \(2\text{.}\)

With a little cleverness, one can come up with \(0/0\) expressions which have a limit of \(\infty\text{,}\) 0, or any other real number. That is why this expression is called indeterminate.

A key concept to understand is that such limits do not really return \(0/0\text{.}\) Rather, keep in mind that we are taking limits. What is really happening is that the numerator is shrinking to \(0\) while the denominator is also shrinking to \(0\text{.}\) The respective rates at which they do this are very important and determine the actual value of the limit.

An indeterminate form indicates that one needs to do more work in order to compute the limit. That work may be algebraic (such as factoring and canceling), it may involve using trigonometric identities or logarithm rules, or it may require a tool such as the Squeeze Theorem. In Section 6.7 we will learn yet another technique called L'Hôpital's Rule that provides another way to handle indeterminate forms.

Some other common indeterminate forms are \(\infty-\infty\text{,}\) \(\infty\cdot 0\text{,}\) \(\infty/\infty\text{,}\) \(0^0\text{,}\) \(\infty^0\) and \(1^{\infty}\text{.}\) Again, keep in mind that these are the “blind” results of directly substituting \(c\) into the expression, and each, in and of itself, has no meaning. The expression \(\infty-\infty\) does not really mean “subtract infinity from infinity.” Rather, it means “One quantity is subtracted from the other, but both are growing without bound.” What is the result? It is possible to get every value between \(-\infty\) and \(\infty\)

Note that \(1/0\) and \(\infty/0\) are not indeterminate forms, though they are not exactly valid mathematical expressions either. In each, the function is growing without bound, indicating that the limit will be \(\infty\text{,}\) \(-\infty\text{,}\) or simply not exist if the left- and right-hand limits do not match.

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 \(\lim\limits_{x\to\infty}f(x)=\infty\) by combining this definition with Definition 1.6.2.

Horizontal asymptotes can take on a variety of forms. Figure 1.6.13.(a) shows that \(f(x) = x/(x^2+1)\) has a horizontal asymptote of \(y=0\text{,}\) where \(0\) is approached from both above and below.

Figure 1.6.13.(b) shows that \(f(x) =x/\sqrt{x^2+1}\) has two horizontal asymptotes; one at \(y=1\) and the other at \(y=-1\text{.}\)

Figure 1.6.13.(c) shows that \(f(x) = \sin(x)/x\) has even more interesting behavior than at just \(x=0\text{;}\) as \(x\) approaches \(\pm\infty\text{,}\) \(f(x)\) approaches \(0\text{,}\) but oscillates as it does this.

We can analytically evaluate limits at infinity for rational functions once we understand \(\lim\limits_{x\to\infty}\frac{1}{x}\text{.}\) As \(x\) gets larger and larger, the \(1/x\) gets smaller and smaller, approaching \(0\text{.}\) We can, in fact, make \(1/x\) as small as we want by choosing a large enough value of \(x\text{.}\) Given \(\varepsilon\text{,}\) we can make \(1/x\lt \varepsilon\) by choosing \(x>1/\varepsilon\text{.}\) Thus we have \(\lim_{x\to\infty} 1/x=0\text{.}\)

It is now not much of a jump to conclude the following: \begin{align*} \lim_{x\to\infty}\frac1{x^n}\amp=0\amp\lim_{x\to-\infty}\frac1{x^n}\amp=0 \end{align*}

Now suppose we need to compute the following limit: \begin{equation*} \lim_{x\to\infty}\frac{x^3+2x+1}{4x^3-2x^2+9}. \end{equation*}

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 denominator. Doing this, we get \begin{align*} \lim_{x\to\infty}\frac{x^3+2x+1}{4x^3-2x^2+9} \amp = \lim_{x\to\infty}\frac{1/x^3}{1/x^3}\cdot\frac{x^3+2x+1}{4x^3-2x^2+9}\\ \amp =\lim_{x\to\infty}\frac{x^3/x^3+2x/x^3+1/x^3}{4x^3/x^3-2x^2/x^3+9/x^3}\\ \amp = \lim_{x\to\infty}\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 \(1/x^n\text{,}\) we see that the limit becomes \begin{equation*} \frac{1+0+0}{4-0+0}=\frac14. \end{equation*}

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 \(b_mx^m/x^n\text{.}\) Since \(n=m\text{,}\) this will leave us with the limit \(a_n/b_m\text{.}\) If \(n\lt m\text{,}\) then after dividing through by \(x^m\text{,}\) all the terms in the numerator will approach \(0\) in the limit, leaving us with \(0/b_m\) or \(0\text{.}\) If \(n>m\text{,}\) and we try dividing through by \(x^m\text{,}\) we end up with the denominator tending to \(b_m\) while the numerator tends to \(\infty\text{.}\)

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

With care, we can quickly evaluate limits at infinity for a large number of functions by considering the long run behaviour using“dominant terms” of \(f(x)\text{.}\) For instance, consider again \(\lim\limits_{x\to\pm\infty}\frac{x}{\sqrt{x^2+1}}\text{,}\) graphed in Figure 1.6.13.(b). The dominant terms are \(x\) in the numerator and \(\sqrt{x^2}\) in the denominator. When \(x\) is very large, \(x^2+1 \approx x^2\text{.}\) Thus \begin{align*} \sqrt{x^2+1}\amp\approx \sqrt{x^2}= \abs{x}\amp\frac{x}{\sqrt{x^2+1}}\amp \approx \frac{x}{\abs{x}}\text{.} \end{align*}

This expression is \(1\) when \(x\) is positive and \(-1\) when \(x\) is negative. Hence we get asymptotes of \(y=1\) and \(y=-1\text{,}\) respectively. We will show this more formally in the next example.