The tangent problem and the recognition of a tangent as the limit of secant lines, instantaneous velocity as the limit of average velocities, the area problem solved by inscribed polygons and rectangles, and the observation that both ancient problems dissolve under the same limiting manoeuvre — the idea Section 2.2 will define.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 2 — Limits
A Preview of Calculus
Objectives
Five outcomes, and none of them is a calculation you will be graded on. They are the reasons the next four sections exist.
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 106-115 — the section these objectives are drawn from
Warm-up
Chapter 1 gave you every function you will need. This chapter asks a question none of those tools can answer.
Discussion prompt
You know the slope between two points is the rise over the run. What happens to that formula if you ask for the slope at a single point on a curve?
Hint: Write the formula down and put the same point in both slots.
Answer:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \;\xrightarrow{\;\text{one point}\;}\; \frac{0}{0} \]
It collapses. With only one point, both the rise and the run are zero, and zero over zero is not a number — it is the absence of an answer.
Yet a curve plainly HAS a direction at each point; a ball plainly has a speed at each instant. The quantity exists, and the algebra of Chapter 1 cannot reach it. That gap is what this whole chapter is about.
Concept
Two problems that resisted algebra for two thousand years share a single resolution. In each, the quantity you want cannot be computed directly, but a sequence of quantities you CAN compute gets arbitrarily close to it. The answer is defined to be what that sequence approaches.
the limiting process — The manoeuvre common to the tangent and area problems: replace an uncomputable quantity by a family of computable approximations, then identify the answer as the single value those approximations approach as the approximation is refined.
\[ \text{answer} = \lim_{\text{refinement}} (\text{approximation}) \]
Notice what is being claimed. Not that some approximation eventually equals the answer — none of them ever does — but that they close in on exactly one number, and that number deserves the name.
Figure (svg): The tangent problem and the area problem side by side, both resolved by the same limiting idea
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 106-107
Section
Section 1
Concept
A secant line through two points on a curve has a slope you can compute by subtraction. A tangent line touches at a single point, and there the slope formula gives zero over zero. The idea is to compute secant slopes with the second point ever nearer, and see what those slopes approach.
secant and tangent lines — A secant line passes through two points of a curve; its slope is the difference in outputs over the difference in inputs. A tangent line touches the curve at one point and has the same direction as the curve there.
\[ m_{\text{sec}} = \frac{f(x) - f(a)}{x - a}, \qquad x \ne a \]
The requirement that x not equal a is doing serious work. The secant slope is undefined at the very point we care about, so the limiting process must approach that point without arriving — which is exactly the shape of the definition in Section 2.2.
Figure (svg): A secant line cutting a curve at two points beside a tangent line touching it at one
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 106-110 — the tangent problem
Picture it
Three secants through the point one comma one on the parabola.
Figure (svg): Three secant lines through a fixed point on a parabola, each drawn to a nearer second point, converging on the tangent
Each secant is a genuine line with a genuine slope, computed by ordinary subtraction. As the second point slides in, the slopes settle toward 2 — a number no single secant achieves.
Worked example
Example 2.1. Ordinary arithmetic, repeated.
\[ \text{For } f(x) = x^2 \text{ at } a = 1, \text{ compute secant slopes for } x = 2, 1.5, 1.1, 1.01. \]
Write the secant slope formula
Why: Difference of outputs over difference of inputs.
\[ m = \frac{x ^{2} - 1}{x - 1} \]
Evaluate at the far point
Why: Four minus 1 over 2 minus 1.
\[ \text{at } x = 2: m = 3 \]
Move the second point nearer
Why: 2.25 minus 1 over 0.5.
\[ \text{at } x = 1.5: m = 2.5 \]
Nearer again
Why: 1.21 minus 1 over 0.1.
\[ \text{at } x = 1.1: m = 2.1 \]
Nearer still
Why: 1.0201 minus 1 over 0.01.
\[ \text{at } x = 1.01: m = 2.01 \]
Figure (svg): A table of secant slopes as the second point approaches from both sides, closing in on two
\[ m_{\text{sec}} \to 2 \text{ as } x \to 1 \]
Verify: factor the secant formula and see why
Why: For x not equal to 1, the numerator factors: x squared minus 1 is x minus 1 times x plus 1, and cancelling leaves simply x plus 1. So every secant slope is x plus 1, which visibly approaches 2 as x approaches 1. That algebraic simplification explains the table rather than merely confirming it, and it is exactly the technique Section 2.3 will systematise. Note the cancellation was legal only because x is not 1 — at x equal to 1 the expression really is 0 over 0.
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 108-109
Matching
For the parabola through the point one comma one.
Match the pairs
Why: Every slope is exactly the second coordinate plus 1, because the secant formula simplifies to x plus 1. The pattern makes the destination obvious: as x closes on 1, the slope closes on 2, and it does so without ever arriving.
Worked example
Checkpoint 2.1. A point and a slope are enough.
\[ \text{Find the tangent line to } f(x) = x^2 \text{ at } (1, 1), \text{ given the slope is } 2. \]
Use point-slope form from Section 1.2
Why: A point on the line and its slope.
\[ y - 1 = 2(x - 1) \]
Distribute
Why: Two x minus 2.
\[ y - 1 = 2 x - 2 \]
Solve for y
Why: Add 1.
\[ y = 2 x - 1 \]
Sanity-check the point lies on it
Why: At x equal to 1 the line gives 1.
\[ \text{passes through } (1, 1) \]
Figure (svg): The solution to Worked example the tangent line's equation shown as a ladder of expressions, one row per legal move
\[ y = 2x - 1 \]
Verify: check the line touches without crossing
Why: Setting x squared equal to 2x minus 1 gives x squared minus 2x plus 1 equal to 0, which is x minus 1 all squared. The repeated root at x equal to 1 says the line meets the parabola at exactly one point, counted twice — which is precisely what touching rather than crossing looks like algebraically. A secant would have produced two distinct roots. That double root is a reliable signature of tangency for a parabola.
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 109-110
Trap
\[ m = \frac{x^2 - 1}{x - 1} \quad \text{at } x = 1 \]
Substitute directly to get the tangent slope
Why: The student puts x equal to 1 into the secant formula.
\[ m = \frac{0}{0} \quad \text{(undefined)} \]
The conclusion drawn is often that the tangent slope does not exist, which is wrong — the parabola plainly has a direction at that point.
\[ m = \frac{x^2-1}{x-1} = x + 1 \quad \text{for } x \ne 1, \;\text{ so }\; m \to 2 \]
Simplify FIRST, while x is still not 1, then let x approach 1
Why: The cancellation is legal at every input except the one we are approaching, and that is enough.
Zero over zero is not a value but a signal: it says the direct substitution has failed and some algebra is required first. Nearly every limit in Section 2.3 begins by producing this form and then removing it, and treating it as a dead end rather than a prompt is the single most costly misreading in the chapter.
Fill the middle
The difference quotient for the squaring function, before letting the gap close.
Fill in the blanks
\fracx + 1___ = \frac______ = ___ \quad (x \ne 1)
Why: Factoring the difference of squares exposes the common factor, and cancelling leaves x plus 1. The cancellation requires x not to be 1, which is exactly the condition the limiting process respects: we approach 1 without ever evaluating there.
Two truths and a lie
All three are about tangents and secants.
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. Tangency is a LOCAL condition: the line matches the curve's direction near the point of contact, and it is free to meet the curve again far away. The tangent to a sine wave at the origin is the line y equals x for the purposes of tangency, yet it crosses the wave repeatedly. Believing tangents touch exactly once causes real trouble at inflection points, where the tangent crosses the curve at the very point it touches.
Prediction
Commit before reasoning.
Predict first
What does getting 0/0 from a secant slope formula tell you?
Correct: Direct substitution failed; algebra is needed first.
\[ \frac{x^2-1}{x-1} \to 2, \qquad \frac{x^3-1}{x-1} \to 3, \qquad \text{both } \tfrac{0}{0} \text{ at } x=1 \]
Why: Zero over zero is called an indeterminate form precisely because it determines nothing: different functions producing it can approach entirely different values, or none. For the squaring function the secants approach 2; for a function with a corner they approach different values from each side and there is no tangent. The form is a prompt to factor, rationalise, or otherwise simplify — and that is the technique the whole of Section 2.3 is built from.
Section
Section 2
Concept
Average velocity over a time interval is the change in position divided by the elapsed time — a secant slope on the position graph. Velocity at a single instant would need a zero elapsed time, so it is again zero over zero, and again it is defined as what the averages approach.
average and instantaneous velocity — Average velocity over an interval is the change in position divided by the length of the interval. Instantaneous velocity at a moment is the value the average velocities approach as the interval shrinks to that moment.
\[ v_{\text{avg}} = \frac{s(t) - s(a)}{t - a}, \qquad v_{\text{inst}} = \lim_{t \to a} v_{\text{avg}} \]
A speedometer reading is an instantaneous velocity, so the quantity is not exotic — it is the one every driver watches. What is remarkable is that a perfectly familiar number could not be defined at all until the limit was available.
Figure (svg): A position graph with average velocity over shrinking intervals shown as secant slopes
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 110-112 — instantaneous velocity
Picture it
A falling body, with average velocities over ever shorter intervals from one second.
Figure (svg): A position graph with average velocity over shrinking intervals shown as secant slopes
Each dashed line is an average velocity over a real interval, and each is a secant slope. As the interval shrinks the secants swing toward the solid tangent, whose slope is the instantaneous velocity, 32 feet per second.
Worked example
Example 2.2. A falling object, four intervals.
\[ \text{A ball falls } s(t) = 16t^2 \text{ feet. Find average velocities on } [1,3], [1,2], [1,1.5], [1,1.1]. \]
Write the average velocity formula
Why: Change in position over change in time.
\[ v = \frac{s(t) - s(1)}{t - 1} \]
Evaluate on the widest interval
Why: 144 minus 16 over 2.
\[ \text{on } [1, 3]: 64 \text{ft} / s \]
Halve the interval
Why: 64 minus 16 over 1.
\[ \text{on } [1, 2]: 48 \text{ft} / s \]
Halve again
Why: 36 minus 16 over 0.5.
\[ \text{on } [1, 1.5]: 40 \text{ft} / s \]
And again
Why: 19.36 minus 16 over 0.1.
\[ \text{on } [1, 1.1]: 33.6 \text{ft} / s \]
Figure (svg): The solution to Worked example average velocities closing in shown as a ladder of expressions, one row per legal move
\[ v_{\text{avg}} \to 32 \text{ ft/s as the interval shrinks to } t = 1 \]
Verify: simplify the quotient algebraically
Why: The average velocity is 16 t squared minus 16 over t minus 1, which factors as 16 times t minus 1 times t plus 1, over t minus 1. Cancelling gives 16 times t plus 1 for every t other than 1, and that expression approaches 32 as t approaches 1. The table's trend is confirmed exactly, and note the cancellation is again licensed by t not being 1 — the same manoeuvre as the tangent problem, because it IS the tangent problem.
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 111-112
Sorting
Ask whether an interval or a moment is involved.
Sort into buckets
Sort each quantity.
Every instantaneous quantity on this list is a tangent slope and every average one is a secant slope. That the speedometer needed a limit to be defined at all is worth sitting with: the number is utterly familiar and its definition is two thousand years younger than the question.
Worked example
Checkpoint 2.2. Sometimes there is no formula, only data.
\[ \text{A runner's position (m) at } t = 1, 1.5, 1.9, 1.99 \text{ s is } 4, 6.5, 8.7, 8.97. \text{ Estimate the velocity at } t = 2, \text{ where } s = 9. \]
Compute the average velocity on each interval to t = 2
Why: Change in position over change in time.
\[ \text{on } [1, 2]: \frac{9 - 4}{1} = 5 \]
Next interval
Why: 9 minus 6.5 over 0.5.
\[ \text{on } [1.5, 2]: 5 m / s \]
Next
Why: 9 minus 8.7 over 0.1.
\[ \text{on } [1.9, 2]: 3 m / s \]
Next
Why: 9 minus 8.97 over 0.01.
\[ \text{on } [1.99, 2]: 3 m / s \]
Read the trend
Why: The last two agree.
\[ \text{about } 3 m / s \]
Figure (svg): The solution to Worked example velocity from a table shown as a ladder of expressions, one row per legal move
\[ v(2) \approx 3 \text{ m/s} \]
Verify: ask whether the estimate is trustworthy
Why: The last two intervals both give 3, which is encouraging — but with data rather than a formula, this is an ESTIMATE and nothing more. The rounding in the position figures limits how far the table can be pushed: at an interval of 0.001 seconds, rounding to two decimals in the positions would swamp the answer entirely. That is an honest limitation of numerical estimation, and it is the reason Section 2.3 develops algebraic techniques rather than relying on tables.
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 112-112
Error analysis
A student computes the average velocity of a falling body on the interval from 1 to 3 seconds.
Annotate
On: \( s(3) - s(1) = 144 - 16 = 128 \;\Longrightarrow\; v_{\text{avg}} = 128 \text{ ft/s} \)
Checking units is the cheapest error detector available in this chapter. A slope is always a change in the output's units divided by a change in the input's, and any answer whose units do not come out that way is wrong before you check the arithmetic.
Fill the middle
The falling body, over the interval from one second to two seconds.
Fill in the blanks
v_48} = \frac______ = \frac______ = ___
Why: The average velocity on that interval is 48 feet per second. Shrinking the interval toward one second sends this toward 32 — the instantaneous velocity there — and the fact that 48 is well above 32 reflects the ball speeding up throughout the interval.
Prediction
Commit before reasoning.
Predict first
For a ball speeding up, is the average velocity on [1, 2] larger or smaller than the instantaneous velocity at t = 1?
Correct: Larger — the interval includes later, faster moments.
\[ v_{\text{avg}}[1,2] = 48 > 32 = v(1) \]
Why: The average over an interval is a blend of every instantaneous velocity in it, and for an accelerating ball those are all at least the velocity at the left endpoint. So the average must exceed the instantaneous value at the start, which matches the numbers: 48 against 32. This reasoning is the seed of the Mean Value Theorem in Section 4.4, which will say that the average over an interval is exactly the instantaneous value at SOME interior point.
Ranking
Average velocities of the falling body on intervals starting at one second.
Put in order
Why: The values are 32, 33.6, 40, 48 and 64. They decrease steadily as the interval shrinks, closing on the instantaneous value from above — which is what an accelerating object must produce. Monotone approach from one side is common but not guaranteed; what the limit demands is only that the values get arbitrarily close.
Section
Section 3
Concept
Geometry gives areas for shapes with straight edges and no others. To find the area of a curved region, inscribe shapes you can measure, use more and more of them, and identify the area as what the totals approach.
the method of exhaustion — Approximating a curved region by inscribed or circumscribed figures with straight edges, then refining the figures indefinitely. The region's area is the value the approximations approach.
\[ A = \lim_{n \to \infty} A_n \]
Archimedes used this to bound pi between two fractions over two thousand years ago, and it is the same reasoning. What he lacked was not the idea but a language for the final step, which is exactly what the limit supplies.
Figure (svg): Inscribed regular polygons with more and more sides, filling out a circle
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 112-115 — the area problem
Picture it
Regular polygons inscribed in a circle of radius one.
Figure (svg): Inscribed regular polygons with more and more sides, filling out a circle
Every polygon has an area that ordinary geometry can compute. None equals the circle's, and the sequence climbs toward pi — the same structure as the secant slopes, with a different quantity.
Worked example
Example 2.3. Right-endpoint rectangles, twice.
\[ \text{Approximate the area under } y = x^2 \text{ on } [0,2] \text{ with 4 and then 8 rectangles.} \]
Find the width of each rectangle
Why: The interval's length divided by the count.
\[ \text{with } 4:\text{ width } 0.5 \]
Take the height at each right endpoint
Why: The function's value there.
\[ \text{heights } 0.25, 1, 2.25, 4 \]
Add width times height
Why: Half of the heights' sum.
\[ 0.5(7.5) = 3.75 \]
Repeat with 8 rectangles of width 0.25
Why: Heights at 0.25 through 2.
\[ \sum 12.75, \times 0.25 \]
Compare
Why: The total has fallen.
\[ 3.1875 \]
Figure (svg): Rectangles under a curve, first four then twelve, approximating the area beneath it
\[ A_4 = 3.75, \qquad A_8 = 3.1875 \]
Verify: check the direction and the destination
Why: Right endpoints on an increasing function overshoot, since each rectangle's top sits above the curve across its whole base. So both estimates should EXCEED the true area, and both should fall as the count rises — and they do. The exact area turns out to be 8 over 3, about 2.667, which both estimates exceed as predicted. Section 5.1 will compute this exactly; here the point is only that the overshoot is shrinking.
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 113-114
Sorting
On an increasing function, the endpoint chosen decides the direction of the error.
Sort into buckets
Sort each rectangle rule, applied to an increasing function.
Having one of each is what produces a BRACKET, and a bracket is far stronger than a single estimate: it bounds the error rather than just suggesting the answer. Archimedes bounded pi exactly this way, with inscribed and circumscribed 96-gons.
Worked example
Checkpoint 2.3. Left endpoints undershoot, which brackets the truth.
\[ \text{Approximate the same area with 4 LEFT-endpoint rectangles, and bracket the true area.} \]
Take the height at each left endpoint
Why: The function's value at the left of each strip.
\[ \text{heights } 0, 0.25, 1, 2.25 \]
Add width times height
Why: Half of the heights' sum.
\[ 0.5(3.5) = 1.75 \]
Note the direction of this error
Why: On an increasing function, left endpoints sit below the curve.
Bracket the true area between the two estimates
Why: Undershoot below, overshoot above.
\[ 1.75 < A < 3.75 \]
Figure (svg): The solution to Worked example bounding the answer from both sides shown as a ladder of expressions, one row per legal move
\[ 1.75 < A < 3.75 \]
Verify: check the true value lies inside the bracket
Why: The exact area is 8 over 3, about 2.667, which does sit between 1.75 and 3.75. The bracket is wide with only four rectangles, and its width is exactly the difference between the two estimates, which is 2. Doubling the rectangle count roughly halves that gap, so the bracket closes — and the fact that it closes to a single number is precisely the claim the limit will make. Bracketing from both sides is stronger evidence than a one-sided estimate, because it bounds the error rather than merely suggesting it is small.
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 114-115
Trap
\[ A_4 = 3.75, \; A_8 = 3.1875, \; A_{16} = 2.9, \; \ldots \]
Wait for some rectangle count to give the exact area
Why: The student looks for the n at which the answer becomes right.
\[ \text{no finite } n \text{ gives } A_n = \tfrac{8}{3} \]
Every finite approximation overshoots, so the search never terminates and the conclusion drawn is often that the area is unknowable.
\[ A = \lim_{n \to \infty} A_n = \tfrac{8}{3} \]
The area is what the totals APPROACH, not a value any of them attains
Why: The limit is a new kind of statement, not a longer computation.
This is the conceptual step the whole chapter rests on, and it is genuinely new. The same structure appeared with the secants: no secant is the tangent. Being comfortable that a quantity can be pinned down exactly by approximations that never reach it is what separates calculus from the algebra of Chapter 1.
Fill the middle
Four right-endpoint rectangles under the squaring function on the interval from zero to two.
Fill in the blanks
A_4 = 0.5(0.25 + 1 + 2.25 + 4) = 0.5(7.5) = 3.75
Why: The total is 3.75, which overshoots the true area of about 2.667 because right endpoints on an increasing function put every rectangle's top above the curve. Doubling the count to eight brings it down to 3.1875, and the overshoot keeps shrinking.
Two truths and a lie
All three are about approximating areas.
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, and it is the misconception this whole idea is designed around. Every finite collection of rectangles has straight tops and the region has a curved one, so some error always remains. The exact area is defined as the limit — the single number the totals approach — not as any total in the sequence. The same is true of the secants and the tangent, and of the polygons and the circle.
Prediction
Commit before reasoning.
Predict first
For the parabola on [0,2], doubling the rectangle count changes the gap between the left and right estimates how?
Correct: It roughly halves.
\[ A_{\text{right}} - A_{\text{left}} = \Delta x\,[f(b) - f(a)] = \Delta x \cdot 4 \]
Why: The gap between the left and right totals is the width of one strip times the total rise of the function across the interval — every interior term cancels between the two sums, leaving only the ends. So halving the width halves the gap. With four rectangles the gap was 2; with eight it is 1; with a thousand it is 0.008. That the gap goes to zero, and does so predictably, is what guarantees the two sequences converge to the SAME number, which is exactly what makes the area well defined.
Section
Section 4
Concept
The tangent problem and the area problem look unrelated: one is about direction at a point, the other about accumulated quantity over an interval. Both were stuck for the same reason, and both yield to the same three-step move.
the two central problems — The tangent problem asks for the slope of a curve at a point and leads to the derivative. The area problem asks for the area under a curve and leads to the integral. Both are solved by taking a limit of computable approximations.
\[ \text{derivative} = \lim \text{(secant slopes)}, \qquad \text{integral} = \lim \text{(rectangle totals)} \]
That the two are not merely analogous but inverse to each other is the Fundamental Theorem of Calculus, proved in Section 5.3. It is genuinely surprising, and it is the reason the subject is one subject rather than two.
Figure (svg): The tangent problem and the area problem side by side, both resolved by the same limiting idea
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 106-115 — the overview of the two problems
Picture it
The two problems set side by side, step for step.
Figure (svg): The tangent problem and the area problem side by side, both resolved by the same limiting idea
Read the two columns across rather than down. Each line of one matches a line of the other, and the fourth line is identical in both — which is the whole reason a single chapter on limits serves the entire subject.
Worked example
Example 2.4. Write each problem in the same three steps.
\[ \text{Express the tangent and area problems in the same form.} \]
State what is wanted and why it is unreachable
Why: Both quantities need something the direct formula cannot supply.
Build a computable approximation
Why: Introduce an extra parameter you can vary.
Refine the approximation
Why: Drive the parameter toward its extreme.
Identify the answer as the limit
Why: The single value approached.
Figure (svg): The solution to Worked example naming the pattern in both shown as a ladder of expressions, one row per legal move
\[ \text{want} \to \text{approximate} \to \text{refine} \to \lim \]
Verify: test the pattern on a third problem
Why: Take the length of a curve. Direct measurement is impossible because rulers are straight; but the curve can be approximated by a chain of straight segments, the segments made shorter, and the length defined as what the chain lengths approach. The same four steps fit exactly — which is a genuine prediction, and Section 6.4 confirms it. A pattern that extends to problems it was not built from is a sign the abstraction is the right one.
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 115-115
Matching
Each quantity is what some sequence approaches.
Match the pairs
Why: Four different quantities, one structure. In every row the left side is something with no direct formula and the right side is a sequence of things with perfectly ordinary formulas. That repetition is why the next section defines the limit once and for all rather than handling each case separately.
Worked example
Checkpoint 2.4. A hint of the theorem that unites them.
\[ \text{A car's velocity is a constant } 60 \text{ mph for } 2 \text{ hours. Relate area and slope here.} \]
Compute the distance travelled
Why: Rate times time.
\[ 120\text{ miles} \]
Draw the velocity graph and find the area under it
Why: A rectangle 2 wide and 60 tall.
\[ \text{area } = 120 \]
Draw the position graph and find its slope
Why: Rising 120 miles over 2 hours.
\[ \text{slope } = 60 \]
State the relationship
Why: Area under the velocity graph is the change in position; slope of the position graph is the velocity.
Figure (svg): The solution to Worked example where the two problems meet shown as a ladder of expressions, one row per legal move
\[ \text{area under } v = \Delta s, \qquad \text{slope of } s = v \]
Verify: check the units confirm the pairing
Why: The area under a velocity graph has units of miles per hour times hours, which is miles — a distance, as claimed. The slope of a position graph has units of miles over hours, which is a velocity. The units alone show that these two operations move between the same two quantities in opposite directions. That is the Fundamental Theorem of Calculus in the simplest possible case, and Section 5.3 proves it holds when the velocity varies.
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 115-115
Error analysis
A student writes about the area under a curve.
Annotate
On: \( A = \lim_{n \to \infty} A_n \approx \tfrac{8}{3} \)
This looks like pedantry and is not. The claim calculus makes is that a sequence of approximations, none exact, pins down an exact answer. Writing 'approximately' throws away the entire achievement and reduces the subject to numerical estimation.
Ranking
The four steps shared by both problems.
Put in order
Why: Step e is the one that is genuinely new. Steps a through d are things a careful Greek geometer did routinely; what was missing for two thousand years was the licence to call the destination the answer, and a precise account of what 'approaches' means. Section 2.5 supplies that account.
Two truths and a lie
All three are about what the chapter is doing.
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 approximations are never exact — no secant is a tangent, no polygon is a circle. What calculus does is show that inexact approximations can pin down an exact answer, provided they close in on one and only one value. Missing this leaves you thinking the subject is elaborate estimation, which is exactly what it is not.
Prediction
Commit before reasoning.
Predict first
Both problems were understood in outline by the ancient Greeks. What did they lack?
Correct: A precise account of what 'approaches a value' means.
\[ \lim_{x \to a} f(x) = L \quad \text{- the statement that took two millennia to make precise} \]
Why: Archimedes computed the approximations superbly and knew what they were closing in on; what he could not do was justify the final step rigorously, so each result needed its own ingenious argument by contradiction. The limit is exactly that missing account, and once stated it turns a collection of individual triumphs into a general method. Section 2.2 gives the working version and Section 2.5 the rigorous one — and the two-thousand-year gap between the problems and their solution is a measure of how hard that definition was to find.
Section
Section 5
Concept
When approximations are given as a table, the answer is read from where the values are heading, not from the final row. Rows can be pushed only so far before rounding and measurement error take over, so the trend is the evidence.
numerical estimation of a limit — Reading the value a sequence of computed approximations is approaching, by examining the trend across several rows rather than the last one. The estimate is only as reliable as the precision of the entries.
\[ \text{tabulate}, \; \text{look for stabilising digits}, \; \text{report those} \]
Approaching from both sides is worth the extra work. Two columns closing on the same value from opposite directions is far stronger evidence than one column drifting, and it also detects the case where the two sides disagree — which Section 2.2 will show is a genuine possibility.
Figure (svg): A table of secant slopes as the second point approaches from both sides, closing in on two
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 108-112 — estimating from tables of values
Picture it
Secant slopes from the left and from the right.
Figure (svg): A table of secant slopes as the second point approaches from both sides, closing in on two
Both columns are heading for 2, and the digits stabilise from the left. The middle of the table is deliberately empty: the value at the point itself is not used and cannot be, since the formula gives zero over zero there.
Worked example
Two columns, one destination.
\[ \text{Estimate what } \frac{x^2-1}{x-1} \text{ approaches as } x \text{ approaches } 1. \]
Tabulate from below
Why: Values at 0.9, 0.99, 0.999.
\[ 1.9, 1.99, 1.999 \]
Tabulate from above
Why: Values at 1.1, 1.01, 1.001.
\[ 2.1, 2.01, 2.001 \]
Compare the two columns
Why: They close on the same value from opposite sides.
\[ \text{both approach } 2 \]
Note the point itself is excluded
Why: The expression is undefined at x equal to 1.
\[ \frac{0}{0}\text{ there} \]
Figure (svg): The solution to Worked example estimating from both sides shown as a ladder of expressions, one row per legal move
\[ \frac{x^2-1}{x-1} \to 2 \text{ as } x \to 1 \]
Verify: confirm algebraically
Why: The expression simplifies to x plus 1 for every x other than 1, and x plus 1 plainly approaches 2. The table and the algebra agree, which is the ideal situation: the table suggests the answer and the algebra proves it. Where the two disagree, trust the algebra — tables can be fooled by values that only look settled, and Section 2.2 gives an example where a table of the wrong inputs suggests a completely wrong answer.
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 108-109
Fill the middle
The secant slopes from below, each row adding another nine.
Fill in the blanks
1.9, \; 1.99, \; 1.999, \; 1.9999, \; \ldots \;\longrightarrow\; 2
Why: The values close on 2 without reaching it, and each row adds one more correct digit. The destination, not the last row, is the answer — and the fact that no row equals 2 is the characteristic feature of a limit rather than a defect of the table.
Worked example
Precision in the inputs limits precision in the answer.
\[ \text{Position data is rounded to 2 decimals. How short an interval is worth using?} \]
Note the uncertainty in each position
Why: Rounding to two decimals means up to 0.005 either way.
\[ \text{position error up to } 0.005 m \]
Note that the difference doubles the worst case
Why: Two rounded values are subtracted.
\[ \text{difference error up to } 0.01 \]
Divide by a short interval
Why: Over 0.01 seconds the error is amplified.
\[ \frac{0.01}{0.01} = 1 m / s\text{ of error} \]
Conclude
Why: The velocity estimate becomes worthless before the interval gets very small.
\[ \text{stop well before } 0.01 s \]
Figure (svg): The solution to Worked example how far to trust a table shown as a ladder of expressions, one row per legal move
\[ \text{error} \approx \frac{2 \times 0.005}{\Delta t} \]
Verify: check the trend of the error
Why: The error is inversely proportional to the interval length, so halving the interval doubles the error. At an interval of 0.1 seconds the error is about 0.1 metres per second, which is tolerable; at 0.001 seconds it is 10 metres per second, which is useless. This is the honest limit of numerical estimation and the practical argument for the algebraic techniques of Section 2.3: algebra has no rounding error, so it can take the interval all the way to zero.
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 112-112
Trap
\[ f(0.9) = 1.9, \; f(0.99) = 1.99, \; f(0.999) = 1.999 \]
Report the final computed value as the answer
Why: The student reads off the bottom row.
\[ \text{the limit is } 1.999 \quad \text{(wrong)} \]
No row of the table is the answer. Every entry is an approximation, and each is closer than the last without any being correct.
\[ 1.9, \; 1.99, \; 1.999, \; \ldots \;\longrightarrow\; 2 \]
Read where the values are HEADING
Why: Look at which digits have stabilised and extrapolate the pattern.
Here the pattern is transparent: each row adds a 9, so the destination is 2. This is the same conceptual point as the rectangles never reaching the exact area — the answer is the destination, not any waypoint. Approaching from both sides makes the destination much easier to see, because two columns converging on one number leave little room for doubt.
Sorting
Ask whether the digits have stabilised and whether the data supports them.
Sort into buckets
Sort each situation.
The one-sided case is weak for a reason that will matter enormously in Section 2.2: a function can approach different values from the left and from the right, and a single column would never reveal it. Tabulating both sides is not thoroughness for its own sake, it is the only way to detect that failure.
Prediction
Commit before reasoning.
Predict first
A table of values suggests a limit of 0. Could the true limit be something else?
Correct: Yes. Badly chosen inputs can suggest an answer that is simply wrong.
\[ \sin\!\left(\tfrac{\pi}{x}\right) = 0 \text{ at } x = 1, \tfrac{1}{2}, \tfrac{1}{3}, \ldots \;\text{ yet the limit does not exist} \]
Why: The sine of pi over x, evaluated at x equal to 1, one half, one third and so on, gives 0 every time — suggesting the limit is 0. In fact the function oscillates between negative 1 and 1 without settling anywhere near the origin, and has no limit at all. The table happened to sample only the zeros. This is why tables build intuition but never establish a limit, and why the algebraic methods of Section 2.3 and the precise definition of Section 2.5 are needed.
Ranking
Weakest first.
Put in order
Why: The jump that matters is from c to d: everything before it is evidence and everything from it on is proof. The chapter's arc runs along this list — Section 2.2 works at levels a to c, Section 2.3 at level d, and Section 2.5 at level e.
Comparison
Fill the blanks. This table is the whole section.
Comparison matrix
| Tangent problem | Area problem | |
|---|---|---|
| What is wanted | slope at one point | area of a curved region |
| Why it fails | slope needs two points, giving 0/0 | geometry only measures straight edges |
| The approximation | a secant through a nearby point | n rectangles under the curve |
| The refinement | slide the second point in | let the number of rectangles grow |
| Leads to | the derivative, Chapter 3 | the integral, Chapter 5 |
The third and fourth rows are identical in structure, and that is the point. One definition of a limit serves both columns, which is why Chapter 2 comes before Chapters 3 and 5 rather than being split between them.
Pattern
Given a quantity that direct computation cannot reach.
Step five is what turns evidence into an answer, and it is available far more often than students expect. The secant slope for the parabola simplified to x plus 1 in one line, and that single line proves what a hundred table rows only suggest.
Stewart, Calculus: Early Transcendentals 8e, §2.1 The Tangent and Velocity Problems §2.1, pp. 78-82
Check
Secant slopes. Ordinary subtraction.
Check your understanding
For f(x) = x^2, find the secant slope between x = 1 and x = 1.1.
Answer: A
Why: 1.21 minus 1 is 0.21, over 0.1, which is 2.1. Equivalently the slope is x plus 1.
Check
Average velocity. Divide by the elapsed time.
Check your understanding
A ball falls s(t) = 16t^2 feet. Find its average velocity on [1, 2].
Answer: A
Why: The position changes from 16 to 64 feet over one second, giving 48 feet per second.
Check
Approximating an area. Direction of the error.
Check your understanding
Right-endpoint rectangles under an increasing function give an estimate that is:
Answer: A
Why: On an increasing function the right endpoint is the strip's highest point, so every rectangle overshoots.
Real world
A phone's GPS records position once per second, and its accelerometer is used to smooth the result. A navigation app must display a live speed.
Discussion prompt
Explain why the app cannot report a truly instantaneous speed, what it actually computes, and why sampling more often does not straightforwardly fix it.
Hint: Speed at an instant needs a zero-length interval; the phone only has samples.
Answer:
The app computes an average velocity over a short interval — the change in position between two samples divided by the time between them. That is a secant slope, and it is the only thing the data supports.
\[ v_{\text{reported}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \approx v(t) \text{, but is never equal to it} \]
Instantaneous speed would require the interval to shrink to nothing, and the phone has no samples in between. What it reports is always a limit's approximation, never the limit.
Sampling more often does not straightforwardly help, for exactly the reason in the last idea: GPS positions carry a few metres of uncertainty, and dividing a fixed position error by a shorter interval amplifies it. Halving the interval doubles the noise in the speed. That is why navigation apps blend GPS with accelerometer data and smooth aggressively — they are managing precisely the trade-off between interval length and rounding error that the worked example computed.
\[ \text{error in } v \approx \frac{2 \times (\text{position error})}{\Delta t} \;\longrightarrow\; \infty \text{ as } \Delta t \to 0 \]
The mathematics has no such problem: an exact formula can take the interval to zero with no error at all, which is the practical advantage of the algebraic methods coming in Section 2.3.
Commit first
Answer, then rate your confidence honestly.
Predict first
What exactly is a tangent line's slope?
Correct: The value the secant slopes approach as the second point approaches the first.
\[ m_{\text{tan}} = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \]
Why: No secant, however close its two points, is the tangent — every one of them has a slightly different slope. Substituting the point directly gives zero over zero, which is why the problem was hard in the first place. And averaging the two sides is not the definition: it happens to give the right answer for smooth functions and the wrong one for a corner, where the two sides genuinely disagree and no tangent exists. The definition is the limit, and only the limit.
Explain it
They insist that if no rectangle sum ever equals the area, then calculus is just a very good approximation.
Discussion prompt
In four sentences or fewer, explain why the limit gives an exact answer even though every approximation is inexact.
Hint: Ask them how many numbers the sequence could possibly be closing in on.
Answer:
Ask them: the totals are 3.75, 3.19, 2.9, 2.8, and they keep shrinking toward something. How many different numbers could a sequence get arbitrarily close to? Exactly one — if it were closing in on two different numbers it would eventually have to be near both at once, which is impossible once you go past half their separation.
So the sequence singles out one number completely, and that number is what we call the area. It is not a good guess at the area; it is the only value consistent with every approximation at once. The approximations are inexact and what they pin down is not.
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 slope formula, write it out and put the same point in both slots — the zero over zero appears immediately. For secant slopes, simplify the difference quotient algebraically and the trend becomes obvious. For the two velocities, ask whether an interval or a moment is named. For the last, remember that a sequence can close in on only one number, and that number is the answer. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Draw the parabola y equals x squared and mark the point one comma one. Draw three secants from that point to second points at x equal to 2, 1.5 and 1.1, compute all three slopes, and write them beside the picture. Then draw the tangent and write its slope. Beneath, tabulate the secant slopes from both sides at 0.9, 0.99, 0.999 and 1.1, 1.01, 1.001, and write one sentence about what the table shows and one about what it cannot show. On the right half of the page, draw the same parabola on the interval from 0 to 2 with four right-endpoint rectangles, compute their total, then sketch eight rectangles and compute that total. Write the two totals, the direction of the error, and one sentence saying why no finite count is exact. At the bottom, write the four-step manoeuvre — want, approximate, refine, take the limit — and beside it fill in what each step is for the tangent problem and for the area problem. In a margin, write the one sentence explaining why a sequence can approach only one number.
If your secant slopes are not exactly the second x-coordinate plus one, recompute: the difference quotient for the squaring function simplifies to x plus 1 everywhere except at the point itself, and that identity is the fastest check on all three.
Recap
Five things, and not one of them is a technique — they are the reasons the rest of the chapter is necessary.
| If you see | Then |
|---|---|
| 0/0 from direct substitution | Simplify first, then let the input approach |
| A slope at a single point | It is a limit of secant slopes |
| A velocity at an instant | It is a limit of average velocities |
| An area under a curve | It is a limit of rectangle totals |
| Right endpoints, increasing function | The estimate is too large |
| A table of approximations | Read the trend, never the last row |
| Two columns agreeing | Reasonable evidence, still not a proof |
Section 2.2 stops gesturing at the idea and defines it. What does it actually mean for a function to approach a value, what can go wrong, and how do the one-sided cases behave — the questions this section has been raising without answering.
OpenStax Calculus Volume 1, §2.1 A Preview of Calculus §2.1, pp. 106-115 — everything on these slides traces back here
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