Infinite Limits & Vertical Asymptotes | AP Calculus 1.14

1. What an infinite limit says

An infinite limit describes outputs that grow without bound in a definite signed direction as the input approaches a finite target. Positive infinity means the outputs eventually exceed any chosen positive bound; negative infinity means they eventually fall below any chosen negative bound.

lim_{x \to a} f (x) = \infty

lim_{x \to a} f (x) = - \infty

“Eventually” matters: the stated behavior must hold sufficiently close to the target along the relevant domain, not merely at a few sampled inputs. Positive infinity is not a very large real number, and negative infinity is not a very small finite value.

The direction attached to the input is separate from the sign of the output. A left-hand approach can lead to positive or negative infinity; the same is true of a right-hand approach.

lim_{x \to a^{+}} f (x) = - \infty

Infinite-limit notation is meaningful

Writing a limit equal to infinity is standard notation for unbounded behavior. It does not assert that a finite real limit exists. When a question asks for the signed infinite behavior, writing only “does not exist” omits useful information.

Review limit notation and one-sided approaches if you need to separate input direction from output behavior.

2. The definition of a vertical asymptote

The vertical line $x = a$ is a vertical asymptote if at least one relevant one-sided limit is positive or negative infinity. The approaches must use inputs in the real domain.

One signed infinite approach is enough to establish a vertical asymptote
Approach	A sufficient infinite-limit statement
From the right, rising	$lim_{x \to a^{+}} f (x) = \infty$
From the right, falling	$lim_{x \to a^{+}} f (x) = - \infty$
From the left, rising	$lim_{x \to a^{-}} f (x) = \infty$
From the left, falling	$lim_{x \to a^{-}} f (x) = - \infty$

A vertical asymptote is a geometric feature; an infinite limit supplies the mathematical description and justification. One side is enough, so a logarithm can have an asymptote at the boundary of its real domain even when there is no approach from the other side.

The value assigned exactly at the target does not control the asymptote. An original rational formula is usually undefined there, but a piecewise function can assign an isolated finite value on the same vertical line. The nearby infinite behavior remains. A dashed guide is not part of the function's graph.

3. Combine one-sided results correctly

Both sides must agree to use one signed two-sided infinity
Left-hand behavior	Right-hand behavior	Two-sided conclusion
Positive infinity	Positive infinity	$lim_{x \to a} f (x) = \infty$
Negative infinity	Negative infinity	$lim_{x \to a} f (x) = - \infty$
Negative infinity	Positive infinity	No common signed two-sided infinite limit. A vertical asymptote still exists.
Positive infinity	Negative infinity	No common signed two-sided infinite limit. A vertical asymptote still exists.
One side infinite	Other side finite or with no real-domain approach	The available infinite side establishes the asymptote; do not invent a two-sided result.

All signed infinite cases fail to give a finite real limit. Avoid using a combined plus/minus infinity symbol as the answer to a particular two-sided problem: state the actual signs separately.

Bounded oscillation, such as sine of a reciprocal near zero, gives no infinite limit. Even an unbounded function does not automatically approach one signed infinity; repeated returns to finite heights can prevent that conclusion. Verify the limiting behavior, not just the presence of very large sampled outputs.

4. Rational functions: factor, cancel, check, determine signs

For a rational function, denominator zeros are candidates, not automatic asymptotes. The memory aid FACD organizes the complete process.

Factor.
Factor numerator and denominator and record every original denominator zero.
Ask what cancels.
Cancel common factors only at allowed inputs. Preserve original exclusions, and check whether a target factor remains in the denominator.
Check the reduced denominator.
For a fully reduced polynomial quotient, a denominator zero with a nonzero numerator gives a pole. A canceled exclusion with a finite remaining limit is a hole.
Determine signs.
Use factor signs on each neighboring interval and show that the magnitude grows without bound. Write each one-sided limit explicitly.

Q (a) = 0, P (a) \neq 0

The displayed condition refers to the reduced numerator and denominator. It is not enough to evaluate a partly simplified expression with another common factor still present.

Cancellation does not always remove the asymptote

\frac{x - 1}{{(x - 1)}^{2}} = \frac{1}{x - 1}, x \neq 1

This expression still has a pole at one. Zero over zero after substitution is an indeterminate form, not proof of a hole. Review Topic 1.13: removable discontinuities and algebraic manipulation of limits.

5. Sign analysis and multiplicity

Near a rational pole, use the sign of each nonzero factor on an interval that contains no other zero or excluded input. A test input helps identify that sign, but a single large numerical result is not a proof of an infinite limit. Combine the sign with a nonzero limiting numerator and a denominator tending to zero.

After complete cancellation, a useful local form is:

f (x) = \frac{R (x)}{{(x - a)}^{m}}

Here the remaining factor is continuous and nonzero at the target, and the denominator exponent is a positive integer. Its parity controls whether the denominator changes sign; the remaining factor controls which infinity occurs.

Apply multiplicity only after cancellation and with a nonzero remaining factor
Remaining power	Sign of remaining factor near target	Left-hand limit	Right-hand limit
Odd	Positive	Negative infinity	Positive infinity
Odd	Negative	Positive infinity	Negative infinity
Even	Positive	Positive infinity	Positive infinity
Even	Negative	Negative infinity	Negative infinity

Odd remaining multiplicity gives opposite signs; even remaining multiplicity gives the same sign. “Even” does not mean automatically positive: a negative numerator can make both sides negative infinity.

6. Worked examples 1–2: a pole and a hole

Example 1: a basic vertical asymptote

f (x) = \frac{1}{x - 2}

The denominator is zero only at two and the numerator remains one. Immediately to the left, the denominator is negative and approaches zero; immediately to the right, it is positive and approaches zero.

lim_{x \to 2^{-}} f (x) = - \infty

lim_{x \to 2^{+}} f (x) = \infty

Conclusion: the line at input two is a vertical asymptote. The sides have opposite signs, so there is no common signed two-sided infinite limit.

The branches are split at input two. The dashed vertical line marks the asymptote; curves leaving the plotting window continue beyond it.

Example 2: distinguish a removable exclusion from a pole

g (x) = \frac{x^{2} - 9}{x^{2} - 5 x + 6}

x^{2} - 9 = (x - 3) (x + 3)

x^{2} - 5 x + 6 = (x - 2) (x - 3)

\frac{x^{2} - 9}{x^{2} - 5 x + 6} = \frac{x + 3}{x - 2}, x \neq 2, x \neq 3

At three, the remaining denominator is one, so the nearby limit is finite:

lim_{x \to 3} g (x) = 6

This is a hole, not a vertical asymptote. At two, the reduced numerator approaches five while the remaining denominator approaches zero with the input's side:

lim_{x \to 2^{-}} g (x) = - \infty

lim_{x \to 2^{+}} g (x) = \infty

Conclusion: the only vertical asymptote is at two; the original graph has a removable hole at $(3, 6)$ .

The original domain excludes both two and three. The dashed guide at two and the open point at

(3, 6)

represent different limit behaviors.

7. Worked examples 3–4: even powers and a logarithm

Example 3: both sides rise to positive infinity

h (x) = \frac{3}{{(x + 1)}^{2}}

The only denominator zero is negative one. The squared denominator is positive on both neighboring sides and tends to zero, while the numerator stays positive.

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

lim_{x \to {- 1}^{+}} h (x) = \infty

lim_{x \to - 1} h (x) = \infty

Conclusion: there is a vertical asymptote at negative one. The two-sided infinity statement is valid, while a finite real limit still does not exist.

The even power keeps the denominator positive on both sides of negative one. Both branches rise beyond the upper plotting window, rather than ending there.

Example 4: a real-domain boundary is enough

f (x) = \ln (x)

The real domain contains only positive inputs. As a positive input approaches zero, its logarithm decreases without bound:

lim_{x \to 0^{+}} \ln (x) = - \infty

One way to interpret this is through the inverse exponential relationship: increasingly negative exponential inputs produce positive outputs closer to zero. There is no real-domain left-hand approach to zero for this function.

Conclusion: the vertical line at zero, the vertical axis, is an asymptote. A right-hand infinite limit alone establishes it; no left-hand value or limit is required.

Only positive inputs belong to the logarithm's real domain. The marked reference point is

(1, 0)

; it is not a domain endpoint.

8. Worked example 5: two vertical asymptotes

p (x) = \frac{2 x}{x^{2} - 4}

x^{2} - 4 = (x - 2) (x + 2)

The original denominator zeros are negative two and two. The numerator is negative four at negative two and positive four at two, so neither candidate cancels. Analyze the neighboring signs at each pole.

Test signs support the factor analysis; magnitude comes from the nonzero numerator over a vanishing denominator
Approach	Nearby test input	Signed quotient at the test input	Limit direction
Left of two	$1.9$	$\frac{3.8}{(- 0.1) (3.9)} < 0$	Negative infinity
Right of two	$2.1$	$\frac{4.2}{(0.1) (4.1)} > 0$	Positive infinity
Left of negative two	$- 2.1$	$\frac{- 4.2}{(- 4.1) (- 0.1)} < 0$	Negative infinity
Right of negative two	$- 1.9$	$\frac{- 3.8}{(- 3.9) (0.1)} > 0$	Positive infinity

The factor signs remain fixed sufficiently close on each side, and the denominator approaches zero while the numerator tends to a nonzero number. Therefore:

lim_{x \to {- 2}^{-}} p (x) = - \infty

lim_{x \to {- 2}^{+}} p (x) = \infty

lim_{x \to 2^{-}} p (x) = - \infty

lim_{x \to 2^{+}} p (x) = \infty

Conclusion: there are exactly two vertical asymptotes, at negative two and two. Splitting the graph at both poles prevents a false connector through an excluded input.

Three separate branches are drawn. At each dashed vertical guide, the left branch falls without bound and the right branch rises without bound.

9. Common functions: state each side precisely

All trigonometric inputs below are in radians, and $n$ denotes an integer. Tangent and cotangent repeat the same side directions at every pole. Secant and cosecant depend on the parity of that integer.

Standard untransformed functions: domains and signed one-sided behavior
Function	Vertical asymptote inputs	Left-hand behavior	Right-hand behavior
$\frac{1}{x}$	$x = 0$	Negative infinity	Positive infinity
$\frac{1}{x^{2}}$	$x = 0$	Positive infinity	Positive infinity
$\ln (x)$	$x = 0$	No real-domain approach	Negative infinity
$\tan (x)$	$x = \frac{π}{2} + n π$	Positive infinity for every integer n	Negative infinity for every integer n
$\cot (x)$	$x = n π$	Negative infinity for every integer n	Positive infinity for every integer n
$\sec (x)$	$x = \frac{π}{2} + n π$	Even n: positive infinity. Odd n: negative infinity.	Even n: negative infinity. Odd n: positive infinity.
$\csc (x)$	$x = n π$	Even n: negative infinity. Odd n: positive infinity.	Even n: positive infinity. Odd n: negative infinity.

For transformed functions, solve for the input that makes the inner expression reach a pole or domain boundary. A negative scale or a reversed inner input can change the side directions; do not transfer signs from this table without checking the transformation.

10. Holes, asymptotes and assigned point values

Classify the nearby behavior rather than an undefined symbol alone
Feature	Removable hole	Vertical asymptote
Limit evidence	A common finite limit exists.	At least one relevant side tends to a signed infinity.
Rational example	$\frac{x^{2} - 4}{x - 2}$ at input two.	$\frac{1}{x - 2}$ at input two.
Graph marker	An open point at the finite limiting height, possibly with a separate filled point.	A dashed vertical guide, supported by actual unbounded behavior.
Point value	Missing or mismatched; assigning the finite limit repairs it.	May be missing or separately assigned; that value does not remove the infinite behavior.
Single-point repair	Possible and unique when the finite common limit is specified.	Impossible with a finite reassignment.

A vertical asymptote is not a literal wall

F (x) = {\begin{cases} \frac{1}{x - 2} & if x \neq 2 \\ 0 & if x = 2 \end{cases}

This function still has the same infinite side limits at two, but its graph also contains the isolated point $(2, 0)$ on the asymptote line. Thus “the graph can never meet a vertical asymptote” is too broad. A continuous finite branch does not pass through the unbounded approach as if nothing happened, and the isolated point does not repair continuity.

See types of discontinuities and removing finite holes for related comparisons.

11. Read graphs and numerical evidence carefully

Follow each side separately: a rising branch means positive infinity, and a falling branch means negative infinity.
Identify the input: a vertical asymptote is a line at a finite input, not a large output value.
Check the domain: an empty side of a logarithm graph may reflect the real domain rather than missing information.
Separate branches: never connect sampled points across a pole or hole.
Interpret clipping: a curve leaving the top or bottom of a plotting window continues beyond the frame; that edge is not an endpoint of the function.
Verify a guide: a dashed line is a visual aid, not proof of an asymptote.

A finite table can suggest a pattern but cannot establish an infinite limit without additional structure. Values may be large because of scale or because a function later oscillates; rational factor analysis or another valid argument supports the conclusion.

Connect formula, graph and numerical evidence using Topic 1.9: multiple representations of limits.

12. Exam explanations, recognition patterns and common mistakes

A strong justification gives the candidate input, valid simplification or domain reasoning, the signed one-sided limits and the asymptote conclusion. Keep the original exclusions visible and use a complete limit statement including the function.

Useful recognition checks

A reduced rational denominator approaching zero with a nonzero numerator indicates a pole; determine its signs.
A finite limit at an original excluded input indicates a hole, not an asymptote.
A logarithm can approach an asymptote from only the side allowed by its domain.
Multiplicity describes the remaining denominator factor after cancellation, with a nonzero local prefactor.

Mistakes to avoid

Calling every original denominator zero an asymptote.
Assuming any canceled factor makes the point removable.
Writing only “undefined” instead of establishing infinite behavior.
Combining opposite infinities into one two-sided result.
Assuming even multiplicity always means positive infinity.
Using a test value from beyond another zero instead of the adjacent sign interval.
Applying rational denominator rules to every possible function.
Treating the asymptote as a literal barrier or ignoring separately defined point values.

Typical questions

Find all vertical asymptotes and justify them with limits.
Show that a specified vertical line is an asymptote.
Determine the behavior from each side of an input.
Sketch the separate branches and asymptote guides.
Explain why an infinite discontinuity cannot be removed by assigning a finite value.

Use the justification requested by the particular question. A memorized acronym supports organization but does not replace the limit argument.

13. Four practice problems with complete signed solutions

All four original problems are retained. Find every original exclusion, distinguish holes from poles and state the signed side limits before opening each solution.

1. A linear denominator

f (x) = \frac{x + 5}{x - 3}

Check the asymptotes and signed limits

lim_{x \to 3^{-}} f (x) = - \infty

lim_{x \to 3^{+}} f (x) = \infty

The numerator tends to eight and the denominator changes from negative to positive through zero. The only vertical asymptote is at three.

2. A fully removable exclusion

g (x) = \frac{x^{2} - 16}{x - 4}

Check the asymptotes and signed limits

\frac{x^{2} - 16}{x - 4} = x + 4, x \neq 4

lim_{x \to 4} g (x) = 8

There is no vertical asymptote. The only original exclusion is a removable hole at $(4, 8)$ . Cancellation does not fill that point in the original graph.

3. Two poles with different numerator signs

h (x) = \frac{2 x + 1}{x^{2} - 1}

Check the asymptotes and signed limits

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

lim_{x \to {- 1}^{+}} h (x) = \infty

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

lim_{x \to 1^{+}} h (x) = \infty

The denominator factors into input minus one and input plus one. Its zeros are negative one and one; the numerator is negative one and three there, respectively, so neither cancels. Both inputs give vertical asymptotes. Analyze all factor signs, including the numerator sign near negative one.

4. Even multiplicity with a negative numerator

p (x) = \frac{x}{{(x + 2)}^{2}}

Check the asymptotes and signed limits

lim_{x \to {- 2}^{-}} p (x) = - \infty

lim_{x \to {- 2}^{+}} p (x) = - \infty

lim_{x \to - 2} p (x) = - \infty

The squared denominator is positive on both sides and tends to zero, while the numerator stays negative near negative two. The only vertical asymptote is at negative two, and both sides fall to negative infinity.

14. Quick reference: location, magnitude, sign

The geometric line is justified by signed infinite limits
Step	What to establish	Common caution
Record the domain	Find original denominator exclusions or another relevant domain boundary.	A missing value alone does not prove an asymptote.
Simplify validly	Factor and cancel nonzero factors, retaining exclusions.	Partial cancellation can still leave a pole.
Show unbounded magnitude	Use a nonzero reduced numerator over a vanishing denominator, or another valid limit argument.	A finite sample does not prove unbounded limiting behavior.
Find each side sign	Use adjacent factor signs and multiplicity.	Even multiplicity may give positive or negative infinity.
State the conclusion	At least one signed infinite side establishes a vertical asymptote.	Use a two-sided signed infinity only when both relevant sides agree.

Infinite limits concern a finite target input here. Limits as the input itself tends to infinity describe a different question: end behavior.

15. Connections and the next topic

This topic connects the language of limits with the geometry of separate branches near a finite input. It also prepares you to distinguish local unbounded behavior from end behavior.

Topic 1.10: vertical asymptotes are associated with infinite discontinuities.
Topic 1.13: a finite limiting height can repair a hole, while signed unbounded behavior cannot be repaired by a finite point assignment.
Topic 1.15: limits at infinity describe end behavior and horizontal asymptotes; the input, rather than just the output, becomes unbounded.
Unit 5 function analysis: combine domain restrictions, asymptotes and other calculus information when interpreting or sketching a function.
AP Calculus BC Unit 6 extension: improper integrals at infinite discontinuities require separate limit and convergence analysis. A vertical asymptote alone does not decide whether such an integral converges.
Modeling: interpret unbounded mathematical outputs alongside a model's domain and physical assumptions.

Previous: Topic 1.13 — Removing discontinuities →Next: Topic 1.15 — Limits at infinity and horizontal asymptotes →All Unit 1 study notes →Unit 1 practice test collection →

For each asymptote candidate, justify the magnitude and sign from the allowed side or sides, then state the vertical line clearly.

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