Connect limits with asymptote behavior

Each side tells its own story

The curve shows \(f(x) = \dfrac{-2}{x + 3}\), beside the dashed line \(x = -3\). The curve is green above the x-axis, where the outputs are positive, and red below it, where they are negative.

Move the two orange points toward \(-3\). At \(x = -3.1\), on the left, the height is 20. At \(x = -2.9\), on the right, the height is \(-20\).

At distance 0.01 the heights are 200 and −200. As the points come closer to the line, the left height grows without bound and the right height falls without bound.

Each part needs its own statement. A one-sided limit gives that statement, because it uses only the inputs on one side of −3.

Four statements, four pictures

The dashed line is \(x = c\), with one part of the curve on each side.

Press the buttons to send a part up or down. The readouts under the figure give the limit from each side.

A part that goes up is green and has the limit ∞ from its side. A part that goes down is red and has the limit −∞ from its side.

The two sides are independent. The line is a vertical asymptote when at least one side has an infinite limit.

Notation

1. The symbol \(x \to c^-\) means that \(x\) approaches \(c\) from the left, with \(x < c\). The symbol \(x \to c^+\) means that \(x\) approaches \(c\) from the right, with \(x > c\).

2. The first statement says that the heights on the right of c grow without bound, so the part goes up. The second says that they fall without bound, so the part goes down.

   \[\begin{gathered} \lim_{x \to c^+} f(x) = \cg{\infty} \\ \lim_{x \to c^+} f(x) = \cr{-\infty} \end{gathered}\]

3. The line \(x = c\) is a vertical asymptote if the limit from at least one side is \(\infty\) or \(-\infty\).

   \[\begin{gathered} \lim_{x \to c^\pm} f(x) = \pm\infty \\ \Rightarrow\ \cy{x = c} \text{ is a vertical asymptote} \end{gathered}\]

4. If both sides have the same infinity, write one two-sided statement with that infinity. If the sides differ, the two-sided limit does not exist.

When it fails

If both one-sided limits are finite, each part comes to a height at an open ring, so the line is not an asymptote. The picture shows a jump, or a hole when the two heights are equal.

If only one side is infinite, the line is still an asymptote. One infinite one-sided limit is enough.

Example 1: one side only

Write the limit statement of \(f(x) = \ln(x - 3)\) at \(x = 3\), and decide whether \(x = 3\) is a vertical asymptote.

1. The logarithm needs \(x - 3 > 0\), so the function has values only for \(x > 3\), on the right side of 3.

2. As \(x\) approaches 3 from the right, \(x - 3\) is small and positive. The logarithm of a small positive number is a large negative number. At \(x = 3.001\) the height is about \(-6.91\).

   \[\lim_{x \to 3^+} \ln(x - 3) = \cr{-\infty}\]

3. One infinite one-sided limit is enough, so \(x = 3\) is a vertical asymptote although the graph has no left side.

Example 2: the two-sided limit

For \(g(x) = \dfrac{8}{x + 4}\), write the one-sided limits at \(x = -4\). Then decide whether the two-sided limit exists and whether \(x = -4\) is a vertical asymptote.

1. Right of \(-4\), \(x + 4\) is small and positive, so \(g\) is large and positive. Left of \(-4\), \(x + 4\) is small and negative, so \(g\) is large and negative.

   \[\begin{gathered} \lim_{x \to -4^+} g(x) = \cg{\infty} \\ \lim_{x \to -4^-} g(x) = \cr{-\infty} \end{gathered}\]

2. A tempting wrong move: Both one-sided limits are infinite, so the two-sided limit is \(\infty\).

   \[\lim_{x \to -4} g(x) = \infty\]

   The left part is red and falls, so its limit is −∞, not ∞. A two-sided limit of ∞ needs both parts to go up.

3. The infinities have different signs, so the two-sided limit does not exist. One infinite one-sided limit is enough, so \(x = -4\) is a vertical asymptote.

Practice

A function has \(\lim_{x \to 1^-} f(x) = \infty\), \(\lim_{x \to 1^+} f(x) = \infty\), \(\lim_{x \to 5^-} f(x) = -\infty\) and \(\lim_{x \to 5^+} f(x) = 2\). List its vertical asymptotes.

Imagine a dashed line at 1 and at 5. For each side, decide whether the part goes up, goes down, or stops at a finite height.

Questions

The graph shows \(v(x) = \dfrac{6}{(x - 1)(x - 4)}\). Green parts lie above the x-axis, and red parts lie below.

1. Which statement about \(x = 1\) is true?

   • \(\displaystyle\lim_{x \to 1^-} v(x) = \infty\)
   • \(\displaystyle\lim_{x \to 1^+} v(x) = \infty\)
   • \(\displaystyle\lim_{x \to 1} v(x) = \infty\)
   • \(\displaystyle\lim_{x \to 1^-} v(x) = -\infty\)
   Answer

   \(\displaystyle\lim_{x \to 1^-} v(x) = \infty\)

2. What does the graph show at \(x = 4\)?

   • A vertical asymptote, and the two-sided limit does not exist
   • A vertical asymptote, and the two-sided limit is \(\infty\)
   • Not an asymptote, because the two sides differ
   Answer

   A vertical asymptote, and the two-sided limit does not exist

3. Find \(\displaystyle\lim_{x \to 3^-} \ln(6 - 2x)\).

   • \(-\infty\)
   • \(\infty\)
   • 0
   • It does not exist, because \(x > 3\) is outside the domain
   Answer

   \(-\infty\)

Summary

You write each side as its own limit, ∞ for a part that goes up and −∞ for a part that goes down.

The statements describe the graph, because an infinite limit means that the heights grow or fall without bound on that side.

One infinite one-sided limit makes the line a vertical asymptote, and the two-sided limit exists only if both sides agree.

This is one of four skills in topic 1.14 of Which & Why: AP Calculus.