Lesson 1.11 · Unit 1 · CED topic 1.14
Connecting infinite limits and vertical asymptotes
So far, a limit that did not exist was simply a failure. Some of those failures are very informative. When a function grows without bound near a point, the direction it runs off in (up or down) tells you a great deal about the graph. In this lesson we extend limit notation to describe unbounded behavior, and connect it to vertical asymptotes.
Learning objective.
- LIM-2.D. Interpret the behavior of functions using limits involving infinity.
Essential knowledge.
- LIM-2.D.1. The concept of a limit can be extended to include infinite limits.
- LIM-2.D.2. Asymptotic and unbounded behavior of functions can be described and explained using limits.
Infinite limits
We write \(\displaystyle\lim_{x\to c}f(x)=\infty\) when the values of \(f(x)\) grow larger without bound as \(x\) approaches \(c\), and \(\displaystyle\lim_{x\to c}f(x)=-\infty\) when they decrease without bound. The same notation works for one-sided limits. Be careful with the meaning: \(\infty\) is not a real number, so a limit that "equals \(\infty\)" is a limit that does not exist as a real number. Writing \(=\infty\) is a way of describing why it does not exist, and it is the preferred way to answer on the exam when the function is unbounded in one direction.
If the two one-sided limits are \(+\infty\) and \(-\infty\), the two-sided limit does not exist and we do not use \(\infty\) for it; we describe each side separately.
Vertical asymptotes
The line \(x=c\) is a vertical asymptote of \(f\) if at least one of the one-sided limits at \(c\) is \(+\infty\) or \(-\infty\). In words, the graph climbs or dives along the line \(x=c\). Vertical asymptotes are an example of how asymptotic and unbounded behavior of functions can be described and explained using limits.
For \(f(x)=\dfrac{1}{x-2}\), as \(x\to2^+\) the denominator is a small positive number, so \(f\to+\infty\); as \(x\to2^-\) the denominator is a small negative number, so \(f\to-\infty\). The line \(x=2\) is a vertical asymptote even though the two-sided limit does not exist.
Finding vertical asymptotes of rational functions
For a rational function in lowest terms, the vertical asymptotes are at the zeros of the denominator. The steps:
- Factor and cancel common factors. A canceled factor leaves a hole (removable discontinuity), not an asymptote.
- Find the remaining zeros of the denominator. At each, the numerator is not zero, so the quotient is (nonzero)/(small), which is unbounded.
- Decide the sign from each side by checking the sign of each factor just to the left and just to the right of \(c\).
Find \(\displaystyle\lim_{x\to2^+}\dfrac{x+1}{x^2-4}\) and \(\displaystyle\lim_{x\to2^-}\dfrac{x+1}{x^2-4}\).
Near \(x=2\) the numerator is close to \(3\gt0\). The denominator \(x^2-4=(x-2)(x+2)\): for \(x\) slightly greater than \(2\), \((x-2)\gt0\) and \((x+2)\gt0\), so the denominator is a small positive number; for \(x\) slightly less than \(2\), it is a small negative number.
\[\lim_{x\to2^+}\frac{x+1}{x^2-4}=+\infty,\qquad \lim_{x\to2^-}\frac{x+1}{x^2-4}=-\infty.\]
The line \(x=2\) is a vertical asymptote. (The same analysis at \(x=-2\) gives \(\displaystyle\lim_{x\to-2^-}=-\infty\) and \(\displaystyle\lim_{x\to-2^+}=+\infty\), because the numerator is near \(-1\).)
Find the discontinuities of \(f(x)=\dfrac{x^2-1}{x^2-3x+2}\) and describe each with limits.
Factor: \(\dfrac{(x-1)(x+1)}{(x-1)(x-2)}\). The factor \(x-1\) cancels, so for \(x\ne1\), \(f(x)=\dfrac{x+1}{x-2}\).
At \(x=1\). \(\displaystyle\lim_{x\to1}f(x)=\dfrac{2}{-1}=-2\): a hole at \((1,-2)\), a removable discontinuity.
At \(x=2\). The numerator is near \(3\), the denominator \(x-2\to0\): \(\displaystyle\lim_{x\to2^+}f(x)=+\infty\) and \(\displaystyle\lim_{x\to2^-}f(x)=-\infty\). The line \(x=2\) is a vertical asymptote.
Find \(\displaystyle\lim_{x\to3}\dfrac{1}{(x-3)^2}\).
The square is positive on both sides of \(3\) and approaches \(0\), so the quotient is a large positive number from both sides. Both one-sided limits are \(+\infty\), so \(\displaystyle\lim_{x\to3}\dfrac{1}{(x-3)^2}=\infty\) (the limit does not exist as a real number, but the unbounded behavior is the same from each side).
Other functions with vertical asymptotes
Rational functions are not the only source. \(\displaystyle\lim_{x\to0^+}\ln x=-\infty\), so \(x=0\) is a vertical asymptote of \(y=\ln x\). And \(\tan x\) has vertical asymptotes at \(x=\dfrac\pi2+k\pi\): \(\displaystyle\lim_{x\to(\pi/2)^-}\tan x=+\infty\) and \(\displaystyle\lim_{x\to(\pi/2)^+}\tan x=-\infty\), because \(\tan x=\dfrac{\sin x}{\cos x}\) with \(\sin\frac\pi2=1\) and \(\cos x\) a small positive then small negative number.
- \(\displaystyle\lim_{x\to c}f(x)=\infty\) describes unbounded growth; the limit does not exist as a real number.
- \(x=c\) is a vertical asymptote if at least one one-sided limit at \(c\) is \(\pm\infty\).
- For rational functions, cancel common factors first: canceled factors are holes; the remaining zeros of the denominator are asymptotes.
- Decide \(+\infty\) or \(-\infty\) from the signs of the factors on each side of \(c\).
- If one side is \(+\infty\) and the other \(-\infty\), describe each side; do not write a two-sided \(\infty\).
- Find \(\displaystyle\lim_{x\to3^-}\dfrac{2}{x-3}\).
Show answer
For \(x\lt3\) the denominator is a small negative number and the numerator is 2, so the limit is \(-\infty\).
- Find \(\displaystyle\lim_{x\to0}\dfrac{1}{x^2}\).
Show answer
The denominator is a small positive number from both sides, so both one-sided limits are \(+\infty\): the limit is \(\infty\) (it does not exist as a real number).
- Find the vertical asymptotes and holes of \(f(x)=\dfrac{x-5}{x^2-25}\).
Show answer
\(f(x)=\dfrac{x-5}{(x-5)(x+5)}=\dfrac1{x+5}\) for \(x\ne5\). There is a hole at \(x=5\) (limit \(\frac1{10}\)) and a vertical asymptote at \(x=-5\).
- Find \(\displaystyle\lim_{x\to1^+}\dfrac{x+2}{x-1}\).
Show answer
The numerator approaches 3 and the denominator is a small positive number, so the limit is \(+\infty\).
- Find \(\displaystyle\lim_{x\to-1^-}\dfrac{x}{x+1}\).
Show answer
The numerator is near \(-1\). For \(x\lt-1\), \(x+1\) is a small negative number, so \(\dfrac{-1}{\text{small negative}}\) is a large positive number: the limit is \(+\infty\).
- Describe the behavior of \(y=\ln x\) near \(x=0\) using a limit, and state the asymptote.
Show answer
\(\displaystyle\lim_{x\to0^+}\ln x=-\infty\), so the line \(x=0\) (the \(y\)-axis) is a vertical asymptote. The function is not defined for \(x\le0\).