AP Calculus AB/BC · Unit 1 · Topic 1.14
Connecting Infinite Limits and Vertical Asymptotes
Use one-sided sign analysis to describe unbounded behavior near a vertical asymptote.
1. Topic Focus
Build the language of limits, connect numerical, graphical, and algebraic representations, and use continuity theorems with verified hypotheses.
This topic: Use one-sided sign analysis to describe unbounded behavior near a vertical asymptote.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
For 1/(x-2), the left side approaches −∞ and the right side approaches +∞.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
1. Infinite Limits Describe Unbounded Behavior
The statement \(\lim\limits_{x\to a^+}f(x)=+\infty\) means that \(f(x)\) can be made larger than any chosen positive bound by taking \(x\) sufficiently close to \(a\) from the right. Infinity is not a real output that the function reaches, so an infinite limit is a description of behavior rather than a finite limit value.
2. The Vertical-Asymptote Connection
The line \(x=a\) is a vertical asymptote of \(f\) when at least one of the one-sided limits at \(a\) is \(+\infty\) or \(-\infty\). The two sides do not need to agree, and the function does not need to be defined at \(a\).
| Left-hand behavior | Right-hand behavior | Two-sided statement | Conclusion |
|---|---|---|---|
| \(+\infty\) | \(+\infty\) | \(\lim\limits_{x\to a}f(x)=+\infty\) | \(x=a\) is a vertical asymptote |
| \(-\infty\) | \(-\infty\) | \(\lim\limits_{x\to a}f(x)=-\infty\) | \(x=a\) is a vertical asymptote |
| \(+\infty\) | \(-\infty\) | The two-sided limit DNE | \(x=a\) is still a vertical asymptote |
| finite or unavailable | \(\pm\infty\) | The two-sided limit may not exist | One unbounded side is sufficient |
3. Separate Sign from Magnitude
Near a possible asymptote, answer two different questions. First, does the denominator's magnitude shrink toward zero while the numerator remains nonzero? If so, the quotient's magnitude grows without bound. Second, what is the sign on each side? A sign chart answers whether the branch rises toward \(+\infty\) or falls toward \(-\infty\).
4. A Reliable Rational-Function Procedure
- Factor the numerator and denominator completely.
- Cancel common factors while preserving the original domain restrictions.
- Locate zeros of the remaining denominator. These are the vertical-asymptote candidates.
- Choose a test input immediately to the left and right of each candidate, or inspect the signs of the factors.
- Combine the local sign with the fact that the denominator magnitude approaches zero.
- Write both one-sided limits before making a two-sided conclusion.
A zero of the original denominator alone is not enough. If its factor cancels completely, the discontinuity is usually a hole rather than a vertical asymptote.
5. Even and Odd Multiplicity
After simplification, suppose the dominant local form is \(C/(x-a)^n\), where \(C\ne0\) near \(a\).
| Power \(n\) | Denominator sign | Branch pattern |
|---|---|---|
| Even | Positive on both sides | Both infinities have the sign of \(C\) |
| Odd | Changes sign at \(a\) | The two infinities have opposite signs |
This pattern is a shortcut only after cancellation and after confirming that the remaining numerator is nonzero at \(a\).
6. Vertical Asymptotes Beyond Rational Functions
A vertical asymptote can occur even without a rational expression. For example, \(\ln(x-a)\to-\infty\) as \(x\to a^+\), and \(1/\sqrt{x-a}\to+\infty\) as \(x\to a^+\). Their real domains exist only to the right of \(a\), but that one unbounded side still establishes the asymptote. Trigonometric functions can also have vertical asymptotes; at \(x=\pi/2\), tangent approaches opposite infinities from the two sides.
7. Connect Formula, Table, and Graph
In a table, inputs should approach the target separately from the left and right. Outputs with rapidly increasing absolute values suggest an infinite limit, while their signs identify the direction. On a graph, a branch becoming nearly vertical is supporting evidence, but the asymptote is the line \(x=a\), not a point on the graph.
8. The Point Value Does Not Control the Asymptote
Changing or adding \(f(a)\) affects one point only. It cannot change the unbounded behavior of nearby outputs, so no assigned real value removes a vertical asymptote. This differs from a removable discontinuity, where a finite two-sided limit identifies one value that fills the hole.
9. Common Reasoning Errors
- Do not write \(f(a)=\infty\); infinity is not a function value.
- Do not conclude that the two-sided limit is \(+\infty\) when the one-sided signs are opposite.
- Do not call every denominator zero a vertical asymptote before simplifying.
- Do not confuse \(x\to a\), which can produce a vertical asymptote, with \(x\to\pm\infty\), which describes end behavior and may produce a horizontal asymptote.
6. Detailed Worked Example and Error Check
Example 1: An even-power denominator. Analyze \(f(x)=\frac{x+2}{(x-1)^2}\) near \(x=1\). The numerator approaches \(3>0\), and the squared denominator approaches zero through positive values on both sides.
Therefore \(\lim\limits_{x\to1}f(x)=+\infty\), and \(x=1\) is a vertical asymptote.
Example 2: An odd-power denominator. For \(g(x)=\frac{x-4}{x+2}\), the numerator stays negative near \(-2\). The denominator is negative to the left and positive to the right.
The two-sided limit does not exist because the directions differ, but \(x=-2\) is still a vertical asymptote.
Example 3: Cancel first, then classify.
At \(x=1\), the canceled factor creates a hole and \(\lim\limits_{x\to1}h(x)=2/9\). At \(x=-2\), the remaining numerator approaches \(-1\) and the squared denominator is positive, so
Thus \(x=-2\) is a vertical asymptote, while \(x=1\) is not.
Example 4: A one-sided domain. Let \(p(x)=\ln(x-3)\). Its domain requires \(x>3\), and
There is no real-domain approach from the left, yet \(x=3\) is a vertical asymptote because the right-hand behavior is unbounded.
Example 5: Opposite trigonometric branches. Near \(x=\pi/2\), \(\sin x\) remains positive while \(\cos x\) changes from positive to negative. Since \(\tan x=\sin x/\cos x\),
The two-sided limit does not exist, and \(x=\pi/2\) is a vertical asymptote.
7. AP Reasoning Routine
Read one-sided behavior first, choose a matching limit procedure, and justify conclusions with definitions or theorem conditions.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
For each problem, determine the requested one-sided limits and justify every vertical-asymptote conclusion.
(a) Analyze \(f(x)=1/(x-5)^2\) as \(x\to5\).
(b) Analyze \(g(x)=-3/(x+1)^3\) as \(x\to-1\).
(c) Classify the discontinuities of \(h(x)=(x-2)/(x^2-4)\) at \(x=2\) and \(x=-2\).
(d) Find both one-sided limits of \(p(x)=(2x+1)/(x-3)^4\) at \(x=3\).
(e) Analyze \(q(x)=1/\sqrt{x-4}\) at \(x=4\) over its real domain.
(f) Analyze \(r(x)=\ln(x+2)\) at the boundary of its domain.
(g) A table shows \(F(1.9)=-48\), \(F(1.99)=-498\), \(F(2.01)=502\), and \(F(2.1)=52\). State the likely one-sided limits at 2 and classify \(x=2\).
(h) A function has \(\lim\limits_{x\to0^-}s(x)=-\infty\), \(\lim\limits_{x\to0^+}s(x)=+\infty\), and \(s(0)=7\). Explain which conclusions are and are not valid.
Check the solution
In part (a), the squared denominator is positive on both sides, so both one-sided limits are \(+\infty\); \(x=5\) is a vertical asymptote. In part (b), the cubic denominator is negative on the left and positive on the right, so the limits are \(+\infty\) and \(-\infty\), respectively; \(x=-1\) is a vertical asymptote and the two-sided limit does not exist. In part (c), cancellation gives \(h(x)=1/(x+2)\) with the original restrictions retained. At 2, the limit is \(1/4\), so there is a hole. At \(-2\), the left-hand limit is \(-\infty\) and the right-hand limit is \(+\infty\), so there is a vertical asymptote. In part (d), the numerator approaches 7 and the fourth-power denominator is positive, so both limits are \(+\infty\) and \(x=3\) is a vertical asymptote. In part (e), only the right-hand approach belongs to the real domain, and \(\lim\limits_{x\to4^+}q(x)=+\infty\); therefore \(x=4\) is a vertical asymptote. In part (f), the domain begins at \(-2\), and \(\lim\limits_{x\to-2^+}\ln(x+2)=-\infty\), so \(x=-2\) is a vertical asymptote. In part (g), the negative outputs decrease without bound from the left and the positive outputs increase without bound from the right, suggesting \(\lim\limits_{x\to2^-}F(x)=-\infty\) and \(\lim\limits_{x\to2^+}F(x)=+\infty\); the two-sided limit does not exist, but \(x=2\) is a vertical asymptote. In part (h), \(x=0\) is a vertical asymptote and the two-sided limit does not exist. The assigned value \(s(0)=7\) neither removes nor changes the asymptote.