Limits at infinity and horizontal asymptotes, the degree rule for rational functions, oblique asymptotes obtained by polynomial division, the end behaviour of polynomials and of the transcendental families, and the complete curve-sketching procedure combining asymptotes with the derivative tests of Section 4.5.
Subject: Calculus I · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Calculus I · Chapter 4 — Applications of Derivatives
Limits at Infinity and Asymptotes
Objectives
Five outcomes. Together with Section 4.5 they let a graph be drawn from the algebra alone.
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 354-380 — the section these objectives are drawn from
Warm-up
Section 1.2 described a polynomial's end behaviour from its degree and leading coefficient, and Section 2.2 met functions whose values grew without bound near a point.
Discussion prompt
The quotient of 3x squared minus 2x by x squared plus 5 is a ratio of two large quantities when x is large. What does it approach, and why is the answer not obvious?
Hint: Both parts grow without bound, so the ratio is indeterminate.
Answer:
Both the numerator and the denominator grow without bound, so the ratio is an indeterminate form of a kind Section 2.3 did not cover — infinity over infinity rather than zero over zero.
\[ \frac{3x^{2}-2x}{x^{2}+5} \;\to\; ? \quad \text{as } x \to \infty \]
The resolution is the same in spirit: rewrite before evaluating. Dividing top and bottom by x squared turns every problematic term into a reciprocal that visibly vanishes, and the answer comes out as 3. That technique is the section's workhorse.
Concept
A limit at infinity describes the outputs as the input runs away without bound. When the outputs settle toward a fixed value, the corresponding horizontal line is an asymptote, and the curve's ends are pinned down.
limit at infinity — The value the outputs approach as the input increases or decreases without bound. If that value is L, the line at height L is a horizontal asymptote of the graph.
\[ \lim_{x \to \infty} f(x) = L \;\Longrightarrow\; y = L \text{ is a horizontal asymptote} \]
This is the last piece of information a sketch needs. Section 4.5 described the middle of a curve; end behaviour describes what happens beyond every window you could draw.
Figure (svg): A curve approaching a horizontal asymptote from both directions
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 354-358
Section
Section 1
Concept
Rather than approaching a point, the input is allowed to grow without bound. If the outputs approach a single value, that value is the limit and its height is a horizontal asymptote.
horizontal asymptote — A horizontal line the graph approaches as the input grows without bound in one or both directions. A curve may cross a horizontal asymptote, and may have a different one at each end.
\[ \lim_{x \to \infty}\frac{1}{x^{n}} = 0 \quad \text{for every } n > 0 \]
Two features surprise people. A curve may cross its horizontal asymptote, sometimes infinitely often; and a function may have different horizontal asymptotes at its two ends, as the arctangent does.
Figure (svg): A curve approaching a horizontal asymptote from both directions
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 354-360 — limits at infinity and horizontal asymptotes
Picture it
A rational function with a single horizontal asymptote.
Figure (svg): A curve approaching a horizontal asymptote from both directions
The curve rises from below on one side and falls from above on the other, closing on the line at both ends without ever reaching it. The asymptote describes the ends and says nothing about the middle.
Worked example
Example 4.33. Everything else is built from these.
\[ \text{Evaluate } \lim_{x \to \infty}\frac{1}{x}, \; \lim_{x \to \infty}\frac{1}{x^{2}}, \; \lim_{x \to -\infty}\frac{1}{x}. \]
Consider the first
Why: A fixed numerator over a growing denominator.
State it
Why: The outputs approach zero.
\[ \lim 0 \]
Consider the square
Why: The denominator grows even faster.
\[ \text{also } 0 \]
Consider the negative direction
Why: The denominator is large and negative.
State it
Why: Still approaching zero.
\[ \lim 0 \]
Figure (svg): The solution to Worked example the basic reciprocal limits shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to \pm\infty}\frac{1}{x^{n}} = 0 \quad (n > 0) \]
Verify: check the sign of the approach in each case
Why: Approaching from the positive side the reciprocal is positive and shrinking, so the curve comes down to the axis from above; from the negative side it is negative and shrinking in magnitude, so it rises to the axis from below. The limit is 0 either way but the approach differs, which matters when sketching. Note that the reciprocal never equals zero, so the axis is approached without being reached — the defining behaviour of an asymptote.
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 355-356
Sorting
Ask what happens as the input runs away.
Sort into buckets
Sort each function by its limit as x grows without bound.
The sine is the important distinction: it has no limit at all rather than an infinite one. Saying a limit is infinity describes a particular failure, and oscillation is a different failure entirely — exactly the distinction Section 2.2 drew.
Worked example
Checkpoint 4.33. A function with two horizontal asymptotes.
\[ \text{Find the horizontal asymptotes of } f(x)=\arctan x. \]
Recall the range
Why: From Section 1.4.
\[ \text{between } -\frac{\pi}{2}\text{ and } \frac{\pi}{2} \]
Consider the positive direction
Why: The tangent grows without bound as its angle nears pi/2.
\[ \arctan x \to \frac{\pi}{2} \]
Consider the negative direction
Why: Symmetrically.
\[ \arctan x \to - \frac{\pi}{2} \]
State both asymptotes
Why: Two different heights.
\[ y = \frac{\pi}{2}\text{ and } y = -\frac{\pi}{2} \]
Figure (svg): The solution to Worked example different asymptotes at the two ends shown as a ladder of expressions, one row per legal move
\[ y = \pm\tfrac{\pi}{2} \]
Verify: confirm the derivative's behaviour agrees
Why: Section 3.7 computed the arctangent's derivative as one over 1 plus x squared, which is positive but tends to 0 as x grows — a function that rises forever while flattening, which is exactly what approaching a horizontal asymptote from below looks like. Two different asymptotes at the two ends is common for bounded increasing functions, and a sketch must show both.
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 357-358
Trap
\[ y = L \text{ is a horizontal asymptote of } f \]
Conclude the graph never touches the line
Why: The student treats the asymptote as a barrier.
\[ f(x) = \frac{\sin x}{x}: \; y = 0 \text{ is an asymptote and } f \text{ crosses it infinitely often} \]
The curve meets the axis at every multiple of pi and still approaches it as the input grows.
\[ \text{an asymptote constrains the ENDS, not the middle} \]
Read the definition as a statement about the limit only
Why: It says the outputs approach L, not that they stay on one side of it.
Vertical asymptotes are different: a curve genuinely cannot cross one, because the function is undefined there. Horizontal ones carry no such restriction, and forgetting this produces sketches that avoid a line the function actually oscillates about.
Fill the middle
The building block every rational limit at infinity is reduced to.
Fill in the blanks
\lim_0\frac______} = ___ \quad \text___ n > 0
Why: Every such reciprocal tends to zero, and the divide-by-the-highest-power technique works by converting a whole rational function into a combination of these.
Two truths and a lie
All three are about horizontal asymptotes.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The quotient of sine by x has the axis as a horizontal asymptote and crosses it at every multiple of pi — infinitely often. Vertical asymptotes cannot be crossed because the function is undefined there, but horizontal ones carry no such restriction and conflating the two produces wrong sketches.
Prediction
Commit before reasoning.
Predict first
Sine has no limit at infinity. Is that the same as having an infinite limit?
Correct: No: sine stays bounded and never settles, which is a different failure from unbounded growth.
\[ |\sin x| \le 1 \text{ always, and yet } \lim_{x \to \infty}\sin x \text{ does not exist} \]
Why: Both are cases where no finite limit exists, but they are distinguishable and the distinction matters for sketching. An unbounded function's graph escapes every horizontal band; sine's stays inside the band from negative 1 to 1 forever while oscillating within it. Section 2.2 drew exactly this distinction for limits at a point, and it carries over unchanged to limits at infinity.
Section
Section 2
Concept
For a quotient of polynomials, the end behaviour is decided by which degree is larger. A larger denominator sends the quotient to zero; equal degrees give the ratio of the leading coefficients; a larger numerator means no horizontal asymptote.
the degree rule — For a rational function: if the numerator's degree is smaller, the horizontal asymptote is the axis; if the degrees are equal, it is the ratio of leading coefficients; if the numerator's is larger, there is no horizontal asymptote.
\[ \lim_{x \to \infty}\frac{a_{n}x^{n}+\cdots}{b_{m}x^{m}+\cdots} = \begin{cases}0 & n<m\\ a_{n}/b_{m} & n=m\\ \pm\infty & n>m\end{cases} \]
The rule follows from the divide-by-the-highest-power technique, and it is worth having both: the rule for speed and the technique for cases the rule does not cover.
Figure (svg): The three cases for a rational function's horizontal asymptote, decided by comparing degrees
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 358-364 — limits at infinity of rational functions
Picture it
Bottom-heavy, balanced and top-heavy quotients.
Figure (svg): Three rational functions illustrating the three degree cases
The first flattens onto the axis, the second onto a horizontal line at height 2, and the third escapes — but along a slanted line, which the next idea makes precise.
Worked example
Example 4.35. The technique behind the rule.
\[ \text{Evaluate } \lim_{x \to \infty}\frac{3x^{2}-2x}{x^{2}+5}. \]
Identify the highest power in the denominator
Why: Degree two.
\[ x ^{2} \]
Divide every term by it
Why: Top and bottom alike.
\[ \frac{3 - \frac{2}{x}}{1 + 5 / x ^{2}} \]
Take the limit of each piece
Why: The reciprocals vanish.
\[ \frac{3 - 0}{1 + 0} \]
Evaluate
Why: The surviving constants.
\[ 3 \]
Compare with the degree rule
Why: Equal degrees.
Figure (svg): The standard technique for a rational limit at infinity: divide by the highest power
\[ \lim_{x \to \infty}\frac{3x^{2}-2x}{x^{2}+5} = 3 \]
Verify: check numerically and against the rule
Why: At x equal to 100 the quotient is about 2.98, and at 1000 about 2.998 — closing on 3. The degree rule predicts the ratio of leading coefficients, 3 over 1, which matches. The technique also explains WHY the rule holds: dividing by the highest power turns every lower-degree term into a vanishing reciprocal, leaving only the leading coefficients.
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 360-361
Matching
Compare the two degrees.
Match the pairs
Why: The rule is decided entirely by comparing degrees, and when they are equal by the ratio of leading coefficients — never by the constant terms, which vanish in the limit. The third has an oblique asymptote instead, which the next idea finds.
Worked example
Checkpoint 4.35. A square root changes the sign.
\[ \text{Evaluate } \lim_{x \to -\infty}\frac{\sqrt{x^{2}+1}}{x}. \]
Note that the root of x squared is the absolute value
Why: From Section 1.4.
\[ \sqrt{x ^{2}} = | x | \]
For negative x, that is negative x
Why: The sign flips.
\[ | x | = -x \]
Divide by x, which is negative
Why: Take the root's factor out carefully.
\[ \sqrt{x ^{2} + 1} / x = -\sqrt{1 + 1 / x ^{2}} \]
Take the limit
Why: The reciprocal vanishes.
\[ -\sqrt{1} = -1 \]
Figure (svg): The solution to Worked example a subtlety in the negative direction shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to -\infty}\frac{\sqrt{x^{2}+1}}{x} = -1 \]
Verify: check numerically and compare the two directions
Why: At x equal to negative 1000 the quotient is about negative 1.0000005, confirming the limit is negative 1. In the positive direction the same expression tends to positive 1, so this function has two different horizontal asymptotes. The sign comes entirely from the root of x squared being the absolute value rather than x itself — the same correction Section 1.4 insisted on, appearing here where it changes an answer's sign.
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 362-363
Error analysis
A student evaluates a limit in the negative direction.
Annotate
On: \( \lim_{x \to -\infty}\frac{\sqrt{x^{2}+1}}{x} = \lim\frac{x\sqrt{1+1/x^{2}}}{x} = 1 \)
The absolute value correction from Section 1.4 is not a formality. Here it flips the sign of the answer, and it is the reason this function has two different horizontal asymptotes rather than one.
Fill the middle
The rational limit from the worked example, after dividing.
Fill in the blanks
\frac3___} \;\longrightarrow\; \frac______ = ___
Why: Every term with x in a denominator vanishes, leaving the ratio of the leading coefficients. That is exactly what the degree rule states for equal degrees.
Sorting
Compare numerator and denominator degrees.
Sort into buckets
Sort each rational function.
The last one exceeds the denominator's degree by exactly one, which is the case that has an OBLIQUE asymptote — unbounded, but following a definite slanted line. The next idea finds it by division.
Prediction
Commit before reasoning.
Predict first
Why does only the leading coefficient matter for a rational limit at infinity?
Correct: Because dividing by the highest power turns every lower term into a vanishing reciprocal.
\[ \frac{-2x}{x^{2}} = -\frac{2}{x} \;\longrightarrow\; 0 \quad \text{exactly} \]
Why: The technique is exact rather than approximate: after dividing, each lower-degree term is a constant over a positive power of x, and Section 4.6's basic limit says every such term goes to zero. So the limit is exactly the ratio of leading coefficients, not approximately. This is the same reasoning as Section 1.2's end behaviour, now made precise by a limit rather than asserted from a picture.
Section
Section 3
Concept
When the numerator's degree exceeds the denominator's by exactly one, polynomial division writes the function as a linear part plus a remainder that vanishes. The linear part is an oblique asymptote.
oblique asymptote — A slanted line the graph approaches as the input grows without bound. It arises when a rational function's numerator exceeds its denominator in degree by exactly one, and it is the quotient of the polynomial division.
\[ \frac{x^{2}+1}{x} = x + \frac{1}{x} \;\Longrightarrow\; y = x \text{ is an oblique asymptote} \]
Division is what makes it visible. Written as a quotient the behaviour is obscure; written as a line plus a dying remainder it is obvious, and the same technique is used again for integration in Chapter 5.
Figure (svg): A rational function with an oblique asymptote, obtained by polynomial division
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 364-368 — oblique asymptotes
Picture it
A rational function with an oblique asymptote.
Figure (svg): A rational function with an oblique asymptote, obtained by polynomial division
Both branches close on the dashed line as the input runs away, one from above and one from below. The function is unbounded and yet its behaviour at the ends is completely described.
Worked example
Example 4.37. Divide and discard the remainder.
\[ \text{Find the asymptotes of } f(x)=\frac{x^{2}+1}{x}. \]
Check the degrees
Why: Numerator exceeds denominator by one.
Divide
Why: Split the fraction termwise.
\[ x + \frac{1}{x} \]
Identify the vanishing part
Why: The reciprocal.
\[ \frac{1}{x} \to 0 \]
Read off the oblique asymptote
Why: What remains.
\[ y = x \]
Find the vertical asymptote too
Why: The denominator vanishes.
\[ x = 0 \]
Figure (svg): A rational function with an oblique asymptote, obtained by polynomial division
\[ y = x \text{ (oblique)}, \quad x = 0 \text{ (vertical)} \]
Verify: check the approach on both sides
Why: At x equal to 100 the function is 100.01, just above the line y equals x; at x equal to negative 100 it is negative 100.01, just below. So the curve approaches from above on the right and below on the left, which the sign of the remainder predicts directly. Both asymptotes are needed for a sketch, and the vertical one comes from the denominator exactly as in Section 2.2.
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 365-366
Fill the middle
The rational function from the worked example, divided termwise.
Fill in the blanks
\frac1/x+1}___ = x + ___
Why: The reciprocal vanishes as x runs away, leaving the line y equals x as the asymptote. Splitting the fraction is the quickest form of the division when the denominator is a single term.
Worked example
Checkpoint 4.37. Long division supplies the quotient.
\[ \text{Find the oblique asymptote of } f(x)=\frac{x^{2}-3x+2}{x-1}. \]
Check the degrees
Why: Two against one.
Divide the polynomials
Why: Long division.
\[ x - 2,\text{ remainder } 0 \]
Interpret a zero remainder
Why: The function IS the line, where defined.
\[ f(x) = x - 2\text{ for } x \ne 1 \]
State the asymptote
Why: The quotient.
\[ y = x - 2 \]
Note the hole
Why: The factor cancels.
\[ a\text{ removable discontinuity at } x = 1 \]
Figure (svg): The solution to Worked example division when the terms do not split shown as a ladder of expressions, one row per legal move
\[ f(x) = x-2 \;(x \ne 1) \]
Verify: check that no vertical asymptote appears
Why: The numerator factors as x minus 1 times x minus 2, so the x minus 1 cancels and the discontinuity at 1 is a removable hole rather than a vertical asymptote — exactly the distinction Section 2.2 drew. This function coincides with its asymptote everywhere it is defined, which is a degenerate but legitimate case: the remainder is zero, so there is nothing left to vanish.
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 366-367
Trap
\[ f(x) = \frac{x^{2}+1}{x}: \; \text{numerator degree exceeds denominator} \]
Conclude the function has no asymptote
Why: The student applies only the horizontal rule.
\[ \text{no horizontal asymptote} \;\Longrightarrow\; \text{no asymptote at all} \quad \text{(wrong)} \]
The function follows the line y equals x arbitrarily closely at both ends, which is a perfectly good asymptote.
\[ \text{degrees differ by one} \;\Longrightarrow\; \text{divide and read the quotient} \]
Check for an OBLIQUE asymptote whenever the numerator's degree is one higher
Why: No horizontal asymptote does not mean no asymptote.
When the numerator's degree exceeds the denominator's by two or more, the quotient is a parabola or higher and the curve follows that instead — sometimes called a curvilinear asymptote. The principle is the same: divide, and whatever survives the remainder's vanishing is the end behaviour.
Sorting
Compare the degrees, then divide if needed.
Sort into buckets
Sort each rational function by its end behaviour.
The last case is worth knowing exists even though it is rarely asked for. The principle generalises: divide, and the polynomial part is whatever the curve follows at its ends.
Ranking
Finding an oblique asymptote.
Put in order
Why: Step b prevents wasted division when no oblique asymptote exists, and step e is the reminder that a rational function usually has vertical asymptotes too — both kinds are needed for a sketch.
Prediction
Commit before reasoning.
Predict first
Why does polynomial division make the oblique asymptote visible?
Correct: Because it separates the function into a polynomial part plus a vanishing remainder.
\[ f = Q + \frac{R}{D}, \quad \deg R < \deg D \;\Longrightarrow\; \frac{R}{D} \to 0 \]
Why: After division the function is written as a polynomial plus a proper fraction, and a proper fraction always tends to zero at infinity by the degree rule. So the polynomial part is exactly what the curve approaches — not approximately, but in the limit. The quotient is linear only when the degrees differ by one; when they differ by more it is a higher polynomial and the curve follows that.
Section
Section 4
Concept
Exponentials outgrow every polynomial, polynomials outgrow every logarithm, and the trigonometric functions do not settle at all. Knowing the ranking resolves most limits at infinity without algebra.
growth ranking — As the input grows without bound, the exponential outgrows every power, every power outgrows the logarithm, and sine and cosine remain bounded without approaching any limit.
\[ \ln x \ll x^{n} \ll e^{x} \quad \text{as } x \to \infty \]
The ranking is more than a convenience. It is what makes a great many limits obvious that would otherwise need L'Hopital's rule, and it is the intuition behind that rule's results.
Figure (svg): The end behaviour of the standard families, compared
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 368-373 — end behaviour of the standard families
Picture it
What each does far from the origin.
Figure (svg): The end behaviour of the standard families, compared
The fourth row is the one that fails differently: sine has no limit rather than an infinite one, and no asymptote of any kind. The others are all settled by the ranking.
Worked example
Example 4.39. The exponential wins.
\[ \text{Evaluate } \lim_{x \to \infty}\frac{x^{100}}{e^{x}}. \]
Note both parts grow without bound
Why: An indeterminate ratio.
Apply the ranking
Why: The exponential outgrows every power.
Conclude
Why: The denominator dominates.
\[ \text{the } \lim\text{ is } 0 \]
Sanity-check the size of the exponent
Why: A hundred is a large power.
Figure (svg): The solution to Worked example a ratio settled by ranking shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to \infty}\frac{x^{100}}{e^{x}} = 0 \]
Verify: check where the exponential actually overtakes
Why: At x equal to 100 the power is 10 to the 200 and the exponential only about 10 to the 43, so the ratio is still enormous — the power is winning decisively. But by x equal to 1000 the exponential is about 10 to the 434 against the power's 10 to the 300, and the ratio has collapsed. The crossover is late, which is exactly why the ranking must be trusted rather than judged from small values. Section 4.8 will prove it with L'Hopital's rule.
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 370-371
Ranking
Slowest first, as the input grows without bound.
Put in order
Why: The logarithm is slowest, every power beats it, larger powers beat smaller ones, and the exponential beats them all. This single ordering settles a large fraction of the limits at infinity you will meet, without any algebra.
Worked example
Checkpoint 4.39. The logarithm loses to every power.
\[ \text{Evaluate } \lim_{x \to \infty}\frac{\ln x}{\sqrt{x}}. \]
Note both grow without bound
Why: Indeterminate again.
Apply the ranking
Why: Every positive power beats the logarithm.
Conclude
Why: The denominator dominates.
\[ \text{the } \lim\text{ is } 0 \]
Note the power is small
Why: One half is still a positive power.
Figure (svg): The solution to Worked example a logarithmic ratio shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to \infty}\frac{\ln x}{\sqrt{x}} = 0 \]
Verify: check with a large value and note how slow it is
Why: At x equal to a million the logarithm is about 13.8 and the root is 1000, so the ratio is about 0.014 — small, but only after a million. At x equal to 100 the ratio is 0.46, still substantial. The logarithm loses to every positive power however small, but it loses slowly, which is why numerical evidence at modest values can be misleading and the ranking is the reliable guide.
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 371-372
Error analysis
A student evaluates a limit at infinity.
Annotate
On: \( \lim_{x \to \infty}x\sin x = \infty \)
An infinite limit means the values eventually exceed every bound and stay there. Oscillating between large positive and large negative values is a different failure, and calling it infinity misdescribes the graph entirely.
Fill the middle
A ratio of a large power to an exponential.
Fill in the blanks
\lim_0\frac___}___} = ___
Why: The exponential outgrows every power, however large the exponent, so the denominator dominates and the ratio tends to zero. The crossover is late but certain.
Sorting
Distinguish unbounded growth from oscillation.
Sort into buckets
Sort each function's behaviour at infinity.
The third and fifth are the pair to compare: both involve sine, and one has no limit while the other has a perfectly good one of 0. The difference is whether sine is multiplied by something growing or divided by it, and the second case is a squeeze from Section 2.3.
Prediction
Commit before reasoning.
Predict first
At x = 100, x^100 vastly exceeds e^x. What does that tell you about the limit?
Correct: Nothing — the crossover is late, and the ranking settles it.
\[ x = 1000: \; x^{100} \approx 10^{300}, \; e^{x} \approx 10^{434} \]
Why: At x equal to 100 the power is about 10 to the 200 and the exponential about 10 to the 43, so the power is winning by 157 orders of magnitude. By x equal to 1000 the exponential has overtaken decisively. Numerical evidence at moderate values is genuinely misleading here, which is why the ranking must be established by argument — Section 4.8's L'Hopital's rule provides it — rather than by inspection of a table.
Section
Section 5
Concept
Combining this section with Section 4.5 gives a complete procedure. Find the domain and intercepts, check symmetry, locate vertical and end asymptotes, then chart both derivatives — and the curve is determined.
the curve-sketching procedure — A checklist combining algebraic features (domain, intercepts, symmetry, asymptotes) with calculus features (increase, decrease, extrema, concavity, inflections) to produce a graph without plotting points.
\[ \text{domain} \to \text{symmetry} \to \text{asymptotes} \to f' \to f'' \]
The order matters because early steps constrain later ones. Knowing the vertical asymptotes tells you where to split the sign charts, and knowing the symmetry halves the work.
Figure (svg): The full curve-sketching checklist, combining this section with Section 4.5
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 373-380 — drawing the graph of a function
Picture it
Six steps in order.
Figure (svg): The full curve-sketching checklist, combining this section with Section 4.5
Steps three and four are this section's contribution; steps five and six are Section 4.5's. Together they leave nothing about the shape undetermined.
Worked example
Example 4.41. Every step of the checklist.
\[ \text{Sketch } f(x)=\frac{x^{2}}{x^{2}-1}. \]
Domain and intercepts
Why: The denominator factors.
\[ x \ne + - 1;\text{ intercept at the origin} \]
Symmetry
Why: Replace x by its negative.
Vertical asymptotes
Why: Where the denominator vanishes.
\[ x = 1\text{ and } x = -1 \]
End behaviour
Why: Equal degrees.
\[ \text{horizontal asymptote } y = 1 \]
First derivative
Why: Quotient rule.
\[ \text{f' } = -2 x / (x ^{2} - 1) ^{2}: \max\text{ at } 0 \]
Second derivative sign
Why: Concave up outside the asymptotes.
Figure (svg): The solution to Worked example a full sketch shown as a ladder of expressions, one row per legal move
\[ \text{VA } x=\pm 1; \; \text{HA } y=1; \; \text{max at } (0,0) \]
Verify: check the pieces against each other
Why: The function is even, so the two vertical asymptotes are symmetric and the only critical point must be on the axis of symmetry — which it is, at the origin. The derivative is negative for positive x, so the middle branch falls away from its maximum at 0 toward negative infinity at x equal to 1. Outside the asymptotes the function exceeds 1 and decreases toward it, approaching the horizontal asymptote from above. Every feature is consistent, and the symmetry halved the work.
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 375-377
Ranking
Sketching a curve from its formula.
Put in order
Why: Symmetry comes early because it can halve every subsequent step. The asymptotes come before the derivative charts because they determine where those charts must be split, since a vertical asymptote always breaks an interval.
Worked example
Checkpoint 4.41. The same checklist, different end behaviour.
\[ \text{Describe the graph of } f(x)=\frac{x^{2}+1}{x}. \]
Domain and intercepts
Why: The denominator vanishes at 0.
\[ x \ne 0;\text{ no intercepts} \]
Symmetry
Why: Replace x by its negative.
Vertical asymptote
Why: Where the denominator vanishes.
\[ x = 0 \]
End behaviour
Why: Divide.
\[ \text{oblique asymptote } y = x \]
First derivative
Why: Split first, then differentiate.
\[ \text{f' } = 1 - 1 / x ^{2}:\text{ extrema at } +- 1 \]
Figure (svg): The solution to Worked example a sketch with an oblique asymptote shown as a ladder of expressions, one row per legal move
\[ \text{VA } x=0; \; \text{oblique } y=x; \; \text{min } (1,2), \text{ max } (-1,-2) \]
Verify: check that the symmetry and the extrema agree
Why: The function is odd, so a minimum at x equal to 1 forces a maximum at negative 1 with the opposite value — and the values are 2 and negative 2 respectively. The minimum value of 2 on the right branch is above the oblique asymptote, which the curve then approaches from above as x grows. Every feature is consistent with the odd symmetry, and noticing it early meant only half the analysis had to be done.
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 377-378
Trap
\[ f(x)=\frac{x^{2}}{x^{2}-1}: \; \text{find the extrema and sketch} \]
Draw only the middle of the curve
Why: The student charts the derivatives and stops.
Without the horizontal asymptote at height 1, the outer branches have no destination and the sketch will show them wandering off arbitrarily.
\[ \text{HA } y = 1: \text{ the outer branches descend toward it from above} \]
Establish the asymptotes before drawing anything
Why: They frame the picture that the derivative information then fills in.
The two kinds of information do different jobs: asymptotes fix the boundaries of the picture, and the derivatives fix the shape within them. Doing the asymptotes first also tells you where to split the sign charts, since a vertical asymptote always breaks an interval.
Fill the middle
A rational function with equal numerator and denominator degrees.
Fill in the blanks
\frac1}___-1} \;\longrightarrow\; ___ \quad \text___ x \to \pm\infty
Why: Equal degrees give the ratio of leading coefficients, which is 1 over 1. So the horizontal asymptote is at height 1, and the outer branches descend toward it.
Sorting
Match each feature to the check that finds it.
Sort into buckets
Sort each graph feature.
The two groups do genuinely different jobs. The algebraic checks frame the picture and the derivative charts fill it in, which is why the checklist does them in that order.
Prediction
Commit before reasoning.
Predict first
What does discovering a function is even save you?
Correct: Half of every subsequent step.
\[ f(-x)=f(x) \;\Longrightarrow\; \text{every feature at } x \text{ has a partner at } -x \]
Why: An even function's graph is the mirror image of its right half, so analysing the positive inputs determines the whole picture — asymptotes, extrema and inflections all come in symmetric pairs. Section 1.1 made exactly this point about plotting, and it applies with more force here because six separate analyses are being halved rather than one. It also provides a check: an even function with an asymmetric answer has an error somewhere.
Comparison
Fill the blanks. Each is found by a different question.
Comparison matrix
| Kind | Where it comes from | Can the curve cross it? |
|---|---|---|
| Vertical | a denominator vanishing without cancelling | no: the function is undefined there |
| Horizontal | a finite limit at infinity | yes, possibly infinitely often |
| Oblique | division, when degrees differ by one | yes |
| None | degrees differing by two or more, or oscillation | not applicable: there is no line to cross |
The crossing column is the one that surprises people. Only a vertical asymptote is a genuine barrier, and it is a barrier because the function has no value there at all.
Pattern
Given a function to sketch.
The asymptotes come before the derivative charts because they determine where the charts must be split. A sign chart that runs straight through a vertical asymptote is wrong.
Stewart, Calculus: Early Transcendentals 8e, §2.6 Limits at Infinity; Horizontal Asymptotes §2.6, pp. 126-139
Check
The degree rule.
Check your understanding
What is the horizontal asymptote of (3x^2 - 2x)/(x^2 + 5)?
Answer: A
Why: The degrees are equal, so the asymptote is the ratio of leading coefficients, 3 over 1.
Check
Oblique asymptotes. No horizontal does not mean none.
Check your understanding
What is the end behaviour of (x^2 + 1)/x?
Answer: A
Why: Dividing gives x + 1/x, and the reciprocal vanishes, leaving the line y = x.
Check
The growth ranking.
Check your understanding
What is the limit of x^100 / e^x as x grows without bound?
Answer: A
Why: The exponential outgrows every power, however large the exponent.
Real world
A drug is infused at a constant rate and cleared by the body in proportion to its concentration. The resulting blood concentration after t hours is twelve milligrams per litre times the quantity one minus a decaying exponential with rate constant three tenths.
Discussion prompt
Find the long-run concentration, say how the derivative behaves, and explain what the asymptote means for dosing.
Hint: This is a limit at infinity, and the growth ranking settles it.
Answer:
\[ \lim_{t \to \infty}12\left(1 - e^{-0.3t}\right) = 12(1 - 0) = 12 \]
The exponential decays to zero, so the concentration approaches a horizontal asymptote at 12 mg/L. This is the steady state: the rate of infusion and the rate of clearance have come into balance.
\[ C'(t) = 3.6e^{-0.3t} > 0, \quad C'' (t) = -1.08e^{-0.3t} < 0 \]
The derivative is always positive and always decreasing, so the concentration rises throughout while levelling off — one of Section 4.5's four shapes, and exactly what approaching a horizontal asymptote from below looks like.
For dosing this matters practically. The concentration never actually reaches 12, so there is no moment at which the steady state is attained. What clinicians use instead is a threshold: at t equal to about 10 hours the concentration is 11.4, within 5 percent of the asymptote, and that is treated as steady state.
The asymptote also caps what this regimen can achieve. If the therapeutic level required is above 12 mg/L, no amount of waiting will reach it — the infusion rate itself must change. Reading an asymptote as a ceiling rather than a target is the practical content of a limit at infinity.
Commit first
Answer, then rate your confidence honestly.
Predict first
A rational function's numerator has degree one more than its denominator. What follows?
Correct: It has an oblique asymptote, found by division.
\[ \frac{x^{2}+1}{x} = x + \frac{1}{x} \;\longrightarrow\; y = x \]
Why: No horizontal asymptote exists, since the function is unbounded — but dividing writes it as a linear quotient plus a proper fraction, and the fraction vanishes at infinity. So the curve follows the line arbitrarily closely at both ends. Concluding no asymptote at all is the standard over-reading of the degree rule, and it produces sketches whose outer branches wander with no destination.
Explain it
They believe a graph can never touch an asymptote.
Discussion prompt
In four sentences or fewer, show them the counterexample and the distinction.
Hint: Use sine over x.
Answer:
Ask them about the quotient of sine by x. Its limit at infinity is 0, so the horizontal axis is an asymptote — and the function equals zero at every multiple of pi, crossing the line infinitely many times.
A horizontal asymptote only says the outputs SETTLE toward that height, not that they stay on one side. A vertical asymptote is different and genuinely cannot be crossed, but only because the function has no value there at all — so the two kinds behave quite differently and the rule they half-remember belongs to one of them.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the degree rule, compare degrees first and only then look at coefficients. For oblique asymptotes, check whether the numerator's degree is exactly one higher before concluding there is none. For roots, remember the root of x squared is the absolute value, which flips sign for negative x. For the checklist, do symmetry early and asymptotes before the derivative charts. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write the degree rule as a three-case table with an example of each, and add a fourth row for the oblique case. Below, take the quotient of 3x squared minus 2x by x squared plus 5, divide top and bottom by x squared, and show every reciprocal vanishing to leave 3. Beside it, take the quotient of x squared plus 1 by x, split it into a line plus a remainder, and sketch the curve with both its asymptotes drawn as dashed lines. In the middle of the page, write the growth ranking from the logarithm to the exponential, and beside it note that sine has no limit rather than an infinite one. In the lower half, work the full checklist for the quotient of x squared by x squared minus 1: domain, symmetry, both vertical asymptotes, the horizontal asymptote, the derivative sign chart split at the asymptotes, and a sketch. In a margin, write which kind of asymptote a curve can cross and why.
If your derivative sign chart for the last function runs straight through x equals 1 without a break, redo it — a vertical asymptote always splits an interval, and a chart that ignores it will claim the function is monotone across a gap it cannot cross.
Recap
Five things, and with Section 4.5 they draw a curve from the algebra alone.
| If you see | Then |
|---|---|
| Denominator degree larger | Horizontal asymptote y = 0 |
| Equal degrees | Asymptote at the leading coefficient ratio |
| Numerator degree one higher | Divide for an oblique asymptote |
| A root and x tending to negative infinity | The root of x squared is minus x |
| An exponential against a power | The exponential wins |
| A bounded oscillation | No limit, rather than an infinite one |
| A vertical asymptote | Split every sign chart there |
Section 4.7 puts the extreme-value machinery to work on word problems. Optimisation is Section 4.3's method applied to a quantity you must first construct from a description, and the construction is where those problems are won or lost.
OpenStax Calculus Volume 1, §4.6 Limits at Infinity and Asymptotes §4.6, pp. 354-380 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Calculus I — $55/session, free consultation.