AP Calculus AB/BC · Unit 1 · Topic 1.15
Connecting Limits at Infinity and Horizontal Asymptotes
Analyze end behavior and identify finite limits at infinity as horizontal asymptotes.
1. Topic Focus
Build the language of limits, connect numerical, graphical, and algebraic representations, and use continuity theorems with verified hypotheses.
This topic: Analyze end behavior and identify finite limits at infinity as horizontal asymptotes.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
For equal-degree rational functions, the end-behavior limit is the ratio of leading coefficients.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
1. Limits at Infinity Describe the Ends of a Graph
The notation \(x\to+\infty\) means that \(x\) increases without bound, while \(x\to-\infty\) means that \(x\) becomes negative with increasingly large magnitude. Infinity is not an input that can be substituted. Instead, a limit at infinity describes a long-run trend.
This statement says that \(f(x)\) can be made arbitrarily close to the finite number \(L\) by taking \(x\) sufficiently large.
2. Finite End Limits Produce Horizontal Asymptotes
The line \(y=L\) is a horizontal asymptote of \(f\) if either
Only one end is required. A function may approach the same horizontal asymptote on both ends, different horizontal asymptotes on the two ends, or a horizontal asymptote on just one end. Always calculate the positive and negative directions separately.
3. An Asymptote Is an End Trend, Not a Barrier
A graph may cross a horizontal asymptote, even repeatedly. The equation \(y=L\) describes what happens far to the left or right; it does not prohibit finite intersections. The difference \(f(x)-L\) is useful: its magnitude measures the vertical distance from the asymptote, and its sign tells whether the graph is above or below it.
4. Why Reciprocal Powers Disappear
For every positive exponent \(p\),
This fact, together with the limit laws, explains most rational-function calculations. Dividing every term by a suitable power of \(x\) rewrites lower-degree terms as reciprocal powers that vanish.
5. Rational Functions: Derive the Degree Cases
Let \(R(x)=P(x)/Q(x)\), with numerator degree \(n\), denominator degree \(m\), and nonzero leading coefficients \(a_n\) and \(b_m\). Divide numerator and denominator by \(x^m\), the highest denominator power.
| Degree comparison | Dominant calculation | Horizontal-asymptote conclusion |
|---|---|---|
| \(n<m\) | Every numerator term becomes a vanishing reciprocal power | Both end limits are 0, so \(y=0\) |
| \(n=m\) | Only the leading coefficients remain | Both end limits are \(a_n/b_m\) |
| \(n>m\) | A positive power of \(x\) remains | No finite horizontal asymptote follows |
The degree table is a shortcut derived from the division, not a replacement for understanding the limit. Simplify common factors first when they obscure the true degrees.
6. Higher Numerator Degree Requires Sign Analysis
If \(n>m\), the function generally has unbounded or polynomial-like end behavior rather than a horizontal asymptote. Use the leading-term quotient
to determine the sign and direction at each end. An odd remaining power changes sign between \(+\infty\) and \(-\infty\); an even remaining power does not.
7. Radical Expressions Require Absolute Value
When a square root contains \(x^2\), extracting the dominant factor gives \(\sqrt{x^2}=|x|\), not \(x\). Therefore
This is why a radical quotient can have different limits on its two ends. At \(+\infty\), \(|x|=x\); at \(-\infty\), \(|x|=-x\).
8. Oscillation Can Still Have a Horizontal Asymptote
Oscillation alone does not force a limit to fail. If its amplitude shrinks, the squeeze theorem can establish a finite end limit. For example, because \(-1\le\sin x\le1\),
The oscillating quotient approaches zero even though it crosses that level many times.
9. Read Tables and Graphs by Direction
A table for \(x\to+\infty\) should use increasingly large positive inputs; a table for \(x\to-\infty\) should use increasingly negative inputs. On a graph, trace the far-right and far-left tails independently. A finite limiting height indicates a horizontal asymptote, while outputs growing without bound do not.
10. Common Reasoning Errors
- Do not substitute an infinity symbol as though it were a number.
- Do not assume the two ends have the same limit.
- Do not claim that a graph cannot cross a horizontal asymptote.
- Do not replace \(\sqrt{x^2}\) with \(x\) when analyzing the negative end.
- Do not conclude that every rational function has a horizontal asymptote.
- Do not confuse \(x\to a\), which studies local behavior, with \(x\to\pm\infty\), which studies end behavior.
6. Detailed Worked Example and Error Check
Example 1: Equal degrees. Evaluate both end limits of
Divide every term by \(x^2\):
Thus \(y=5/2\) is a horizontal asymptote on both ends.
Example 2: Denominator degree is larger.
The graph approaches the horizontal asymptote \(y=0\) as \(x\to+\infty\) and as \(x\to-\infty\).
Example 3: The numerator degree is larger. For
the dominant quotient is \(2x^3/x^2=2x\). Hence
There is no horizontal asymptote. The degree comparison identifies the lack of a finite end limit; the leading terms determine the signs.
Example 4: A radical creates different end limits. Analyze
Factor \(x^2\) from the radical:
Therefore
The same function has horizontal asymptote \(y=1\) on the right and \(y=-1\) on the left.
Example 5: Shrinking oscillation. Let \(p(x)=2+\sin x/x\). Since \(|\sin x|\le1\),
By the squeeze theorem,
The line \(y=2\) is a horizontal asymptote, and the graph crosses it whenever \(\sin x=0\) with \(x\ne0\).
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.
Find both end limits and identify every horizontal asymptote. Show the algebra or theorem supporting each conclusion.
(a) \(f(x)=(4x^2-x)/(2x^2+7)\).
(b) \(g(x)=(3x-1)/(x^2+5)\).
(c) \(h(x)=(2x^3+1)/(x^2-4)\).
(d) \(p(x)=(5x+2)/\sqrt{4x^2+9}\).
(e) \(q(x)=\sqrt{9x^2+1}/x\).
(f) \(r(x)=4+\cos x/x\), for \(x\ne0\).
(g) A table shows \(F(-1000)\approx-1.002\), \(F(-100)\approx-1.020\), \(F(100)\approx2.970\), and \(F(1000)\approx2.997\). State the likely end limits and horizontal asymptotes.
(h) Explain why a solution of \(r(x)=4\) does not contradict the horizontal asymptote found in part (f).
Check the solution
In part (a), equal degrees give both limits \(4/2=2\), so \(y=2\) is a horizontal asymptote on both ends. In part (b), the denominator degree is larger, so both limits are 0 and \(y=0\) is the horizontal asymptote. In part (c), the leading quotient is \(2x\); the limits are \(+\infty\) and \(-\infty\) at the positive and negative ends, respectively, so there is no horizontal asymptote. In part (d), dividing through by \(|x|\) gives \((5x/|x|+2/|x|)/\sqrt{4+9/x^2}\). The limits are \(5/2\) as \(x\to+\infty\) and \(-5/2\) as \(x\to-\infty\), producing two horizontal asymptotes. In part (e), \(q(x)=(|x|/x)\sqrt{9+1/x^2}\), so the limits are 3 and \(-3\); the asymptotes are \(y=3\) on the right and \(y=-3\) on the left. In part (f), \(|\cos x/x|\le1/|x|\), so both limits are 4 by the squeeze theorem and \(y=4\) is the horizontal asymptote. In part (g), the data suggest \(\lim\limits_{x\to-\infty}F(x)=-1\) and \(\lim\limits_{x\to+\infty}F(x)=3\), so the horizontal asymptotes are \(y=-1\) on the left and \(y=3\) on the right. In part (h), a horizontal asymptote describes a limiting trend as \(|x|\) grows; it is not a barrier. Indeed, \(r(x)=4\) whenever \(\cos x=0\) and \(x\ne0\), so the graph may cross the asymptote repeatedly.