Limits at Infinity & Horizontal Asymptotes | AP Calculus 1.15

1. Limits at infinity describe end behavior

A limit at infinity describes what happens to a function's outputs as the input becomes unbounded in one direction. The input does not approach a particular real number. Instead, we examine the far-right or far-left behavior along the function's domain.

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

This says the outputs eventually stay arbitrarily close to the finite number represented by the limiting height as the input increases without bound.

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

The second statement describes the far-left end. The two end limits can agree, be different, be infinite, or fail to exist because the outputs keep oscillating. Evaluate each available direction separately.

Infinity describes a direction, not a real input

You cannot substitute infinity as though it were an ordinary number. Rewrite the expression using valid algebra and known limiting behavior. Infinite-limit notation is meaningful for unbounded outputs, but it does not give a finite limiting height.

Compare this with Topic 1.14: infinite limits near a finite input. Review limit notation for the distinction between the approaching input and the resulting output.

2. Connect a finite end limit to a horizontal asymptote

The line $y = L$ is a horizontal asymptote if the function approaches that finite height at either end of its domain.

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

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

One of these statements is enough; both may hold. A limit is the analytic description, while the horizontal line is the geometric feature. If an end limit is positive or negative infinity, that particular end does not give a horizontal asymptote. Check the other end before concluding there is none.

A real-valued function can have at most two distinct horizontal asymptotes: one finite limiting height for each end.
If both ends approach the same height, count one distinct line.
If neither end has a finite limiting height, there is no horizontal asymptote.
A function can cross a horizontal asymptote finitely or infinitely many times, even arbitrarily far out, while the distance from the line tends to zero.

A finite graphing window can suggest a limiting height, but it cannot prove end behavior. An apparent flattening may be temporary. Support the graph with a limit argument.

3. The three rational-function degree rules

These rules apply to a quotient of polynomials with a nonzero denominator polynomial. Let the numerator and denominator degrees be represented by $n$ and $m$ , with nonzero leading coefficients $a$ and $b$ .

\frac{P (x)}{Q (x)} \sim \frac{a x^{n}}{b x^{m}} = \frac{a}{b} x^{n - m}

The comparison symbol means their ratio tends to one at either end; it is an end-behavior statement, not an identity for every input.

Rational functions: compare polynomial degrees before calculating
Degree comparison	End-limit result	Horizontal asymptote
$n < m$	Both ends tend to zero.	$y = 0$
$n = m$	Both ends tend to the ratio of leading coefficients.	$y = \frac{a}{b}$
$n > m$	Unbounded magnitude at both ends; determine each sign from the leading coefficient and power.	None at either end.

When the numerator degree is larger, an odd difference in degrees gives opposite signs at the two ends; an even difference gives the same sign, controlled by the leading-coefficient ratio. Do not write an ambiguous plus/minus infinity as the answer to a specific end limit.

If the degree difference is one, polynomial division gives a slant asymptote. For a larger difference it gives a higher-degree polynomial asymptote. In either case the remainder term tends to zero after division. Finite holes and poles do not change a polynomial quotient's far-end degree comparison, but retain its original domain exclusions.

4. Divide by the highest denominator power

Identify the highest power in the denominator.
Divide every term in both numerator and denominator by that same power. This is valid away from zero, which is enough for end behavior.
Reciprocal positive integer powers tend to zero at either end. Keep any remaining positive powers and determine their signs.
Use the limit laws only where their hypotheses hold. A finite nonzero limiting denominator permits the quotient law.
Write the limit at each end and then name any finite horizontal line.

lim_{x \to \infty} \frac{1}{x^{m}} = 0

The same conclusion holds at negative infinity when the exponent is a positive integer. For general real powers, the real domain may restrict the directions available.

Review limit laws and algebraic manipulation. A quotient whose numerator and denominator both grow without bound is an indeterminate form, not automatically one or zero.

5. Worked examples 1–3: all three degree cases

Example 1: lower numerator degree

f (x) = \frac{5 x + 3}{2 x^{2} - x + 1}

The numerator has degree one and the denominator degree two. Divide every term by the square of the input:

\frac{5 x + 3}{2 x^{2} - x + 1} = \frac{\frac{5}{x} + \frac{3}{x^{2}}}{2 - \frac{1}{x} + \frac{1}{x^{2}}}

Every reciprocal term tends to zero. The numerator tends to zero, while the denominator tends to two, a nonzero value.

lim_{x \to - \infty} \frac{5 x + 3}{2 x^{2} - x + 1} = 0

lim_{x \to \infty} \frac{5 x + 3}{2 x^{2} - x + 1} = 0

The horizontal asymptote is $y = 0$ at both ends. This denominator has no real zeros, so the original function is defined for all real inputs.

Example 2: equal degrees

f (x) = \frac{3 x^{2} - 4 x + 7}{6 x^{2} + 5}

Both degrees are two. The leading coefficients are three and six. Dividing by the square of the input verifies the coefficient rule:

\frac{3 x^{2} - 4 x + 7}{6 x^{2} + 5} = \frac{3 - \frac{4}{x} + \frac{7}{x^{2}}}{6 + \frac{5}{x^{2}}}

\frac{3}{6} = \frac{1}{2}

lim_{x \to - \infty} \frac{3 x^{2} - 4 x + 7}{6 x^{2} + 5} = \frac{1}{2}

lim_{x \to \infty} \frac{3 x^{2} - 4 x + 7}{6 x^{2} + 5} = \frac{1}{2}

The horizontal asymptote is $y = \frac{1}{2}$ . The function crosses it at $(\frac{9}{8}, \frac{1}{2})$ ; crossing does not contradict an end limit.

The dashed guide marks the common limiting height. The marked crossing is exact; the visible edges are plotting boundaries, not endpoints of the domain.

Example 3: higher numerator degree and a slant asymptote

f (x) = \frac{2 x^{3} + x - 1}{x^{2} + 5}

\frac{2 x^{3} + x - 1}{x^{2} + 5} = \frac{2 x + \frac{1}{x} - \frac{1}{x^{2}}}{1 + \frac{5}{x^{2}}}

The denominator tends to one. The linear term in the numerator determines the unbounded sign at each end.

lim_{x \to - \infty} \frac{2 x^{3} + x - 1}{x^{2} + 5} = - \infty

lim_{x \to \infty} \frac{2 x^{3} + x - 1}{x^{2} + 5} = \infty

There is no horizontal asymptote. Polynomial division gives more information:

\frac{2 x^{3} + x - 1}{x^{2} + 5} = 2 x + \frac{- 9 x - 1}{x^{2} + 5}

The remainder quotient tends to zero at both ends, so the slant asymptote is $y = 2 x$ . The function's difference from that line tends to zero.

6. Worked examples 4–5: dominant growth

Example 4: a nonconstant polynomial

f (x) = 2 x^{3} + 5 x - 1

The cubic term dominates the linear and constant terms. Factor out that leading term to justify the comparison:

2 x^{3} + 5 x - 1 = 2 x^{3} (1 + \frac{5}{2 x^{2}} - \frac{1}{2 x^{3}})

The parenthesized factor tends to one. The positive-coefficient odd power tends to negative infinity on the left and positive infinity on the right.

lim_{x \to - \infty} 2 x^{3} + 5 x - 1 = - \infty

lim_{x \to \infty} 2 x^{3} + 5 x - 1 = \infty

A nonconstant polynomial has no horizontal asymptote. A constant polynomial is the exception: its graph is already the horizontal line approached at both ends.

Example 5: an exponential denominator

f (x) = \frac{3 x - 1}{e^{x}}

At positive infinity, the exponential grows faster than the fixed linear numerator. Splitting the quotient separates two standard limits:

\frac{3 x - 1}{e^{x}} = 3 \frac{x}{e^{x}} - \frac{1}{e^{x}}

lim_{x \to - \infty} \frac{3 x - 1}{e^{x}} = - \infty

lim_{x \to \infty} \frac{3 x - 1}{e^{x}} = 0

On the left, the numerator is negative with unbounded magnitude and the positive denominator tends to zero, so the quotient tends to negative infinity. The line $y = 0$ is a horizontal asymptote at the right end only. A failure to have a finite limit at one end does not cancel a horizontal asymptote at the other.

7. Worked example 6: different left and right asymptotes

Example 6: a radical requires an absolute value

f (x) = \frac{x}{\sqrt{x^{2} + 1}}

The original expression is defined for all real inputs because the expression inside the square root is positive. For nonzero inputs, factor the square from inside the radical:

\sqrt{x^{2} + 1} = \sqrt{x^{2} (1 + \frac{1}{x^{2}})} = | x | \sqrt{1 + \frac{1}{x^{2}}}

\frac{x}{\sqrt{x^{2} + 1}} = \frac{x}{| x |} \frac{1}{\sqrt{1 + \frac{1}{x^{2}}}}

At positive inputs the sign factor is one. At negative inputs it is negative one. The remaining positive radical tends to one at either end.

\frac{x}{\sqrt{x^{2} + 1}} = \frac{1}{\sqrt{1 + \frac{1}{x^{2}}}}, x > 0

\frac{x}{\sqrt{x^{2} + 1}} = \frac{- 1}{\sqrt{1 + \frac{1}{x^{2}}}}, x < 0

lim_{x \to - \infty} \frac{x}{\sqrt{x^{2} + 1}} = - 1

lim_{x \to \infty} \frac{x}{\sqrt{x^{2} + 1}} = 1

There are two distinct horizontal asymptotes: $y = - 1$ on the left and $y = 1$ on the right. Replacing the square root of the input squared with the input itself would give the wrong left-end answer.

Separate end limits give two horizontal lines. Neither line has to be reached at a finite input for it to be an asymptote.

8. Compare relative growth with limits

Growth comparisons concern sufficiently large positive inputs and fixed parameters. They are not claims about every input. A root is itself a power: the square root has exponent one half. The relative order of positive powers depends on their exponents.

lim_{x \to \infty} \frac{x^{p}}{b^{x}} = 0, p > 0, b > 1

lim_{x \to \infty} \frac{\ln (x)}{x^{p}} = 0, p > 0

Fixed exponents: larger powers eventually dominate smaller powers
Power comparison at positive infinity	Quotient limit
$p < q$	$lim_{x \to \infty} \frac{x^{p}}{x^{q}} = 0$
$p = q$	$lim_{x \to \infty} \frac{x^{p}}{x^{q}} = 1$
$p > q$	$lim_{x \to \infty} \frac{x^{p}}{x^{q}} = \infty$

For example, a logarithm grows more slowly than a positive square root, a square root more slowly than a linear power, and a fixed-base growing exponential more quickly than every fixed positive power. Saying “an exponential denominator gives zero” needs a numerator that grows more slowly and an appropriate input direction.

Optional sequence connection: factorials of positive integers eventually dominate a fixed positive-base exponential. This is a comparison along integer inputs, not an extra real-variable limit technique needed for Topic 1.15. Infinite sequences and series are studied in AP Calculus BC Unit 10.

Later, L’Hôpital’s Rule can evaluate certain indeterminate quotients after its differentiability and limit hypotheses are checked. It is not required for these algebraic examples, and an indeterminate form alone is not a conclusion.

9. Oscillation, squeeze bounds and crossing asymptotes

Sine and cosine remain bounded but do not approach a single value as the input tends to either infinity. Thus neither has a horizontal asymptote. Boundedness alone does not establish convergence.

Dividing a bounded oscillation by a quantity whose magnitude grows without bound can shrink the oscillation to zero. For nonzero inputs:

0 \leq | \frac{\sin (x)}{x} | \leq \frac{1}{| x |}

lim_{x \to - \infty} \frac{\sin (x)}{x} = 0

lim_{x \to \infty} \frac{\sin (x)}{x} = 0

The outer bound tends to zero in either direction, so the Squeeze Theorem gives the two limits. The absolute value avoids reversing an inequality incorrectly at negative inputs.

The horizontal asymptote is $y = 0$ . The function crosses that line at nonzero integer multiples of pi, infinitely often in both directions, while its oscillation amplitude shrinks. The original quotient is undefined at zero; its finite limit there is one, so the graph has a hole unless a value is separately assigned.

Angles are in radians. The open point preserves the original excluded input zero. A horizontal asymptote permits repeated crossings arbitrarily far from the origin.

10. Vertical and horizontal asymptotes compared

Local behavior near an input versus limiting heights at the ends
Feature	Vertical asymptote	Horizontal asymptote
Line	$x = a$	$y = L$
Input behavior	Approaches a finite target from an allowed side.	Becomes unbounded to the left or right.
Output behavior	Unbounded in at least one signed one-sided limit.	Approaches a finite height at at least one end.
Number possible	A function may have many, even infinitely many.	At most two distinct finite heights.
Meeting the line	Nearby unbounded behavior defines it; a separately assigned isolated point can lie on the line.	Crossings can occur finitely or infinitely often.
How to justify	Establish a signed infinite one-sided limit.	Establish a finite limit at an available end.

A zero denominator is only a candidate for a vertical asymptote. A finite limiting height after cancellation may instead indicate a hole. See removing discontinuities and vertical-asymptote reasoning.

11. Four retained practice problems with complete solutions

For each original problem, evaluate the left and right end limits separately, state all horizontal asymptotes and justify the method before opening the solution.

1. Equal linear degrees

f (x) = \frac{4 x - 7}{2 x + 3}

Check the two end limits and asymptotes

lim_{x \to - \infty} \frac{4 x - 7}{2 x + 3} = 2

lim_{x \to \infty} \frac{4 x - 7}{2 x + 3} = 2

Both degrees are one. The leading-coefficient ratio is two, so $y = 2$ is the horizontal asymptote at both ends. The original input exclusion is negative three halves.

2. A larger denominator degree

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

Check the two end limits and asymptotes

lim_{x \to - \infty} \frac{x^{2} + 1}{3 x^{3} - 2} = 0

lim_{x \to \infty} \frac{x^{2} + 1}{3 x^{3} - 2} = 0

The denominator degree is three and the numerator degree two. Dividing by the cubic power leaves a numerator tending to zero and a denominator tending to three. The horizontal asymptote is $y = 0$ at both ends.

3. A larger numerator degree

f (x) = \frac{5 x^{3} + 2 x}{x^{2} - 4}

Check the two end limits and asymptotes

lim_{x \to - \infty} \frac{5 x^{3} + 2 x}{x^{2} - 4} = - \infty

lim_{x \to \infty} \frac{5 x^{3} + 2 x}{x^{2} - 4} = \infty

The leading behavior is five times the input. There is no horizontal asymptote. Long division gives $\frac{5 x^{3} + 2 x}{x^{2} - 4} = 5 x + \frac{22 x}{x^{2} - 4}$ , so the slant asymptote is $y = 5 x$ . The original exclusions are negative two and two.

4. An exponential with different end behaviors

f (x) = \frac{2^{x}}{x^{2}}

Check the two end limits and asymptotes

lim_{x \to - \infty} \frac{2^{x}}{x^{2}} = 0

lim_{x \to \infty} \frac{2^{x}}{x^{2}} = \infty

On the right the exponential dominates the quadratic. On the left the numerator tends to zero while the positive denominator grows without bound, so the quotient tends to zero. Therefore $y = 0$ is a horizontal asymptote at the LEFT end. Checking only the right end would incorrectly miss it. The original input zero remains excluded.

Practice 4 has a left horizontal asymptote even though its right-end limit is infinite. The dotted vertical guide marks the separate pole at zero; branches are never joined across it.

12. Exam strategies, recognition patterns and common mistakes

A reliable answer structure

Identify the available end or ends and the original domain.
Choose a method: degrees, normalization, a root sign split, growth comparison or a squeeze bound.
Show the essential algebra or inequality and state each limit clearly.
Connect every finite end limit to its horizontal line; distinguish slant or unbounded behavior.

Quick recognition patterns

For a rational function, a lower numerator degree gives zero at both ends.
Equal polynomial degrees give the leading-coefficient ratio, such as $\frac{3}{5}$ when the leading coefficients are three and five.
An exponential denominator dominates a fixed polynomial numerator at positive infinity when its base exceeds one. Recheck the left end independently.
A square root of a squared input introduces an absolute value.
Bounded oscillation with shrinking amplitude can have a finite end limit; persistent-amplitude oscillation need not.

Errors to avoid

Using degree rules on functions that are not polynomial quotients.
Dividing only selected terms by the highest power.
Using constant terms instead of leading coefficients for equal polynomial degrees.
Treating infinity as a number or an indeterminate quotient as an answer.
Forgetting the negative-input sign of an absolute value.
Concluding there is no horizontal asymptote after checking only one end.
Assuming an asymptote can never be crossed or reached.
Taking a few table entries or a flat-looking graph as proof of a limit.
Using a later calculus rule without checking its hypotheses.

Typical AP-style tasks

Find and justify all horizontal asymptotes; interpret a limit statement as end behavior; connect a formula with a graph or table; compare the relative magnitudes of two functions; or explain why two ends approach different heights. Use the representation and justification requested by the problem. For practice connecting evidence, review multiple representations of limits.

13. Quick reference and later connections

Choose the method from the expression, then justify the end limit
Function family	Method	Caution
Rational polynomial quotient	Compare degrees; normalize or divide.	State signed limits separately if the quotient grows without bound.
Polynomial	Use the leading term and parity.	Only constant polynomials have finite end limits.
Radical quotient	Factor inside the root; retain absolute values.	The far-left and far-right signs may differ.
Polynomial and exponential	Compare growth in the specified direction.	A conclusion at positive infinity may reverse at negative infinity.
Oscillation	Find a shrinking absolute bound or show persistent variation.	Boundedness alone does not imply a limit.

Connections: Unit 5 uses end behavior alongside derivatives to analyze functions. The BC extension on improper integrals in Unit 6 uses limits to test convergence over unbounded intervals; a horizontal asymptote alone does not decide convergence. BC Unit 10 extends limit reasoning to sequences and series, while BC Unit 9 applies calculus to parametric and polar representations. These are later connections, not additional prerequisites for the examples here.

Modeling connection: a finite end limit may represent a long-term trend, carrying capacity or steady-state level when the model and its units support that interpretation. Approaching a level does not mean reaching it at a finite time, and a mathematical model should not be extended beyond its stated domain without justification.

Previous: Topic 1.14 — Infinite limits and vertical asymptotes →All Unit 1 study notes →Unit 1 practice test collection →

End behavior is a claim about arbitrarily distant inputs. Establish the limit first, then interpret its finite height as a horizontal asymptote.

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