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A First Calculus Unit: Limits, Change, and Accumulation

Build the connection from nearby function values to instantaneous rates, then from rates to total change.

Introductory single-variable calculus · Original self-checks with answers you can reveal

Goals and prerequisites

For first-time calculus learners who can manipulate algebraic expressions, or learners reviewing the meaning behind the rules.

Four focused chapters introduce limits, derivatives, the chain rule, and definite integrals with polynomial examples. This is not a complete Calculus I or AP syllabus. Trigonometric, exponential, and logarithmic derivatives, optimization, integration techniques, and series require further study.

Bring these skills

  • Expand a square, factor a difference of squares, and cancel a nonzero common factor.
  • Evaluate a function, read slope as output change per input change, and state a function’s domain.
  • For a full calculus course beyond this unit, also review trigonometry, exponentials, and logarithms.

What you will be able to do

  • Distinguish a function’s value at a point from its limit as the input approaches that point.
  • Derive a simple derivative from a difference quotient and interpret its sign and units.
  • Differentiate a composite polynomial, write a tangent line, and distinguish net change from total distance.

Begin with D1 and D2. D3–D5 sample ideas taught later, so unfamiliarity is expected on a first pass. Work through chapters in order if calculus is new; use the weak-area links if reviewing. Keep a written record of units, domain restrictions, and the reason for each limit or derivative step. Attempt answers before revealing them.

Start with the diagnostic

D1 checks essential algebra; D2 introduces the first calculus idea. The remaining questions help returning learners choose what to review. These are not prerequisites you must already pass to start.

Question 1

D1 · Algebra readiness: simplify for h ≠ 0.

(x+h)2−x2h\frac{(x+h)^2-x^2}{h}
Answer, reasoning and error check for question 1

Answer: 2x + h

Expand (x + h)² = x² + 2xh + h². Subtract x², factor h, then cancel it using h ≠ 0.

Common mistake: h comes from omitting the cross term 2xh. If this is difficult, review algebraic expansion before Chapter 2.

Need a review? Go to the matching chapter

Question 2

D2 · Limit versus value: find the limit. Is the original quotient defined at x = 1?

lim⁡x→1x2−1x−1\lim_{x\to1}\frac{x^2-1}{x-1}
Answer, reasoning and error check for question 2

Answer: The limit is 2; the original quotient is undefined at x = 1.

For x ≠ 1, factor and cancel to obtain x + 1. Nearby outputs approach 2 from both sides, even though the original rule has no value at 1.

Common mistake: “0/0, so the limit is zero” is invalid. 0/0 is not a number; it tells you direct substitution has not settled the limit.

Need a review? Go to the matching chapter

Question 3

D3 · Meaning of a derivative: a tank’s volume V is measured in liters and time in minutes. What do V(2) = 7 and V′(2) = −3 mean?

Answer, reasoning and error check for question 3

Answer: At 2 minutes the volume is 7 liters and is decreasing instantaneously at 3 liters per minute.

The function value measures an amount. Its derivative measures the local rate of change of that amount, so the derivative has units liters/minute.

Common mistake: “The tank contains −3 liters” confuses a rate with an amount. A negative derivative does not require a negative function value.

Need a review? Go to the matching chapter

Question 4

D4 · Composite functions: differentiate with respect to x.

y=(2x+1)3y=(2x+1)^3
Answer, reasoning and error check for question 4

Answer: y′ = 6(2x + 1)²

The outer cube contributes 3(2x + 1)² and the inner function contributes a factor of 2.

Common mistake: 3(2x + 1)² differentiates only the outside. The input to the outside changes twice as fast as x, so multiply by 2.

Need a review? Go to the matching chapter

Question 5

D5 · Accumulation: a volume changes at a constant −2 liters/minute for 3 minutes. What is its net change?

Answer, reasoning and error check for question 5

Answer: −6 liters, meaning 6 liters were lost.

Rate × duration gives change: (−2 liters/minute) × (3 minutes) = −6 liters.

Common mistake: −6 liters/minute keeps the rate units after multiplying by time. Minutes cancel. Also, this gives a change, not the final volume without an initial value.

Need a review? Go to the matching chapter

Choose your starting point

The algebra or the hole is the obstacle

D1 or D2 was uncertain.

Begin with Chapter 1. If expansion or factoring is still the obstacle, use the algebra unit linked below before tackling difference quotients.

Go to this chapter

You can compute but cannot interpret

D3 was wrong or lacked units.

Use Chapter 2 to separate position, average rate, and instantaneous rate.

Go to this chapter

Nested expressions lose a factor

D4 missed the inner derivative.

Label the inner and outer functions in Chapter 3 before differentiating.

Go to this chapter

Rates and total changes are mixed up

D5 had the wrong sign, units, or interpretation.

Use Chapter 4 to distinguish signed accumulation, final amount, and total distance.

Go to this chapter

Chapter 1

1. Ask what happens nearby, not only at the point

Goal: Use algebra and one-sided behavior to decide a finite two-sided limit.

The statement “f(x) approaches L as x approaches a” describes outputs for inputs arbitrarily close to a, with x different from a. It does not require f(a) to exist. More precisely, any prescribed positive error in the output can be met by taking every allowed input sufficiently close to a on the relevant side or sides.

A two-sided limit exists only when the left-hand and right-hand limits both exist and agree. A table suggests a trend but does not prove it; a graph can also hide a jump or a hole. Use an expression, a definition, or a valid limit rule to justify the conclusion.

Polynomials are continuous, so their limits can be found by substitution. A quotient of continuous functions also permits substitution where its denominator is nonzero. If substitution produces 0/0, simplify or find another argument; do not treat 0/0 as the answer.

Worked example

A removable hole in a graph

Find the limit and explain what happens to the graph at x = 3.

lim⁡x→3x2−9x−3\lim_{x\to3}\frac{x^2-9}{x-3}

  1. Substitution gives 0/0, which does not determine the limit. Factor the difference of squares.

    x2−9=(x−3)(x+3)x^2-9=(x-3)(x+3)

  2. For nearby x other than 3, cancel a nonzero common factor. This equality is restricted to x ≠ 3.

    (x−3)(x+3)x−3=x+3(x≠3)\frac{(x-3)(x+3)}{x-3}=x+3\qquad(x\ne3)

  3. The nearby values agree with a continuous line. Evaluate its limit by substitution.

    lim⁡x→3(x+3)=6\lim_{x\to3}(x+3)=6

The limit is 6. The original graph is the line y = x + 3 with an open point at (3, 6). Assigning a new value f(3) = 6 would fill that hole and make a continuous extension, but the original quotient remains undefined at 3.

Approach the hole from both sides

Approach the hole from both sidesThe line y = x + 3 is drawn with an open circle at (3, 6). Dashed guides meet x = 3 and y = 6. An arrow from smaller x and another from larger x both point toward the hole. The left and right limits equal 6, while f(3) is undefined. The displayed window is 0 ≤ x ≤ 5; the function is defined for all real x except 3.369012345f(x)xy = x + 3 (x ≠ 3)(3, 6)
The quotient (x² − 9)/(x − 3) follows y = x + 3 only for x ≠ 3. Both arrows approach the open point (3, 6), so the limit is 6. The open point means f(3) is undefined; a limit does not fill the hole.

Avoid this trap: A hole and a jump are different. For |x|/x, inputs below 0 give −1 and inputs above 0 give 1. The two-sided limit at 0 does not exist, even though both one-sided limits do.

Practice on your own

Question 1

Find the limit and state whether the original expression is defined at x = 5.

lim⁡x→5x2−25x−5\lim_{x\to5}\frac{x^2-25}{x-5}
Answer, reasoning and error check for question 1

Answer: Limit 10; the original expression is undefined at 5.

Factor x² − 25 = (x − 5)(x + 5). For x ≠ 5, the quotient is x + 5, whose limit is 10.

Common mistake: “No limit because the denominator is zero” confuses the value at the point with behavior nearby.

Need a review? Go to the matching chapter

Question 2

Let f(x) = 2x for x < 1 and f(x) = x + 2 for x ≥ 1. Find the left-hand limit, right-hand limit, f(1), and the two-sided limit at 1.

Answer, reasoning and error check for question 2

Answer: Left limit 2; right limit 3; f(1) = 3; the two-sided limit does not exist.

Use the formula for each side separately. The value f(1) comes from the second branch, but it cannot make the two unequal one-sided limits agree.

Common mistake: 3 uses only the right-hand branch. A two-sided limit must account for inputs approaching from both directions.

Need a review? Go to the matching chapter

Before moving on: Before continuing, explain why a quotient can have a limit where it has no value, and why two different one-sided limits cannot form one two-sided limit.

Chapter 2

2. Turn an average rate into an instantaneous rate

Goal: Build a derivative from its limit definition and keep rate units distinct from amount units.

The average rate of change of f from a to a + h is [f(a + h) − f(a)]/h, where h ≠ 0. It is the slope of the secant line through two points. Its units are output units divided by input units.

The derivative f′(a) is the limit of that quotient as h approaches 0, if a finite limit exists. It is the tangent slope and instantaneous rate. You do not substitute h = 0 into the original quotient; you analyze its nearby behavior.

The derivative also gives a local estimate: for small h, f(a + h) ≈ f(a) + f′(a)h. Multiply the rate at a by the input change, then add that estimated output change to the starting value. This is a tangent-line approximation; without further information, it gives no guaranteed error bound for a particular step size.

A function can be continuous and still have no derivative at a corner. At 0, |x| has left slope −1 and right slope 1, so there is no single derivative there. Differentiability is a stronger requirement than continuity.

Worked example

Position is not velocity

An object’s position is s(t) = t² − 4t + 7 meters, with time in seconds and t ≥ 0. Find its velocity at t = 3 using the definition. Compare it with its average velocity from t = 1 to t = 3.

  1. At t = 3, the position is 4 meters. Build the difference quotient around this point.

    s(3+h)−s(3)h=(3+h)2−4(3+h)+7−4h\frac{s(3+h)-s(3)}{h}=\frac{(3+h)^2-4(3+h)+7-4}{h}

  2. Expand, combine, and cancel h only while h ≠ 0.

    =2h+h2h=2+h(h≠0)=\frac{2h+h^2}{h}=2+h\qquad(h\ne0)

  3. Take the limit to obtain instantaneous velocity, measured in meters per second.

    s′(3)=lim⁡h→0(2+h)=2s^{\prime}(3)=\lim_{h\to0}(2+h)=2

  4. For the average over a fixed interval, use two positions instead. Here s(1) = s(3) = 4.

    s(3)−s(1)3−1=4−42=0\frac{s(3)-s(1)}{3-1}=\frac{4-4}{2}=0

The instantaneous velocity at 3 seconds is +2 m/s, while the average velocity from 1 to 3 seconds is 0 m/s. The object can return to the same position over an interval while moving at its endpoint.

Avoid this trap: f′(a) = 0 does not by itself prove a maximum or minimum. For f(x) = x³, the derivative at 0 is zero, but the function increases through 0. A flat tangent is not automatically a turning point.

Practice on your own

Question 1

Use the definition to find f′(2), where f(x) = x² + 1.

Answer, reasoning and error check for question 1

Answer: f′(2) = 4

The quotient is [(2 + h)² + 1 − 5]/h = (4h + h²)/h = 4 + h for h ≠ 0. Its limit is 4.

Common mistake: 5 is f(2), the point’s height, not f′(2), the tangent slope.

Need a review? Go to the matching chapter

Question 2

A differentiable temperature function T has T(10) = 18 °C and T′(10) = −0.4 °C/min. Estimate T(10.5) with a tangent-line approximation and explain why it is an estimate.

Answer, reasoning and error check for question 2

Answer: Approximately 17.8 °C.

A small input change Δt = 0.5 gives approximate output change T′(10)Δt = −0.2 °C. Add it to 18. The derivative is local, and the actual cooling rate may change during the half minute.

Common mistake: 17.6 °C uses a full minute rather than half a minute. Claiming exactly 17.8 °C also assumes a constant rate that was not given.

Need a review? Go to the matching chapter

Before moving on: Explain a derivative as a limit, a slope, and a rate with units. Then explain why average velocity can be zero while instantaneous velocity is nonzero.

Chapter 3

3. Track every layer of a changing input

Goal: Differentiate a polynomial composition and use the result to build a tangent line.

For a positive integer n, the derivative of xⁿ is nxⁿ⁻¹. Constants differentiate to 0; sums differentiate term by term; a constant multiplier stays in front. These rules are consequences of the limit definition and save repeating the expansion each time.

For a composition y = F(g(x)), changes pass through two stages. If g is differentiable at x and F is differentiable at g(x), the chain rule gives y′ = F′(g(x))g′(x). Evaluate the outer derivative at the inner output, then multiply by the inner derivative.

A tangent line at x = a needs both a point and a slope. The point is (a, f(a)); the slope is f′(a). Its equation is y − f(a) = f′(a)(x − a). It agrees with the function at a and models local change, but need not equal the function elsewhere.

Worked example

Differentiate a fourth power, then find its tangent

Find the derivative of f(x) = (3x − 2)⁴ and the tangent line at x = 1.

  1. Name the inner expression u = 3x − 2. Differentiate the outside with respect to u and the inside with respect to x.

    dduu4=4u3,dudx=3\frac{d}{du}u^4=4u^3,\qquad\frac{du}{dx}=3

  2. Multiply the two rates and replace u with its full expression.

    f′(x)=4(3x−2)3⋅3=12(3x−2)3f^{\prime}(x)=4(3x-2)^3\cdot3=12(3x-2)^3

  3. Find both the point and the slope at x = 1.

    f(1)=1,f′(1)=12f(1)=1,\qquad f^{\prime}(1)=12

  4. Use point-slope form, keeping the tangent’s y distinct from the original formula.

    y−1=12(x−1)⇒y=12x−11y-1=12(x-1)\quad\Rightarrow\quad y=12x-11

The derivative is 12(3x − 2)³, and the tangent line is y = 12x − 11. The line passes through (1, 1) and has slope 12; checking only one of these conditions is not enough.

Avoid this trap: A sum, product, quotient, and composition require different structures. For example, the derivative of x·x is 2x, not 1·1. Do not extend the term-by-term sum rule to multiplication; study the product and quotient rules next.

Practice on your own

Question 1

Differentiate and identify the inner derivative.

y=(x2+1)3y=(x^2+1)^3
Answer, reasoning and error check for question 1

Answer: y′ = 6x(x² + 1)²; the inner derivative is 2x.

The outer derivative is 3u² and u = x² + 1 has derivative 2x. Multiply them: 3(x² + 1)² × 2x.

Common mistake: 3(x² + 1)² omits the varying inner rate. 6x³ differentiates parts of the expression without respecting its composition.

Need a review? Go to the matching chapter

Question 2

Find the tangent line to f(x) = x³ − 2x at x = 2.

Answer, reasoning and error check for question 2

Answer: y − 4 = 10(x − 2), or y = 10x − 16

f(2) = 8 − 4 = 4 and f′(x) = 3x² − 2, so f′(2) = 10. A line with that slope through (2, 4) has the stated equation.

Common mistake: y = 10x + 4 uses f(2) as a y-intercept. The known point is at x = 2, not x = 0.

Need a review? Go to the matching chapter

Before moving on: Mark the inner and outer functions before each chain-rule calculation. For a tangent line, verify both that it passes through the point and that its slope equals the derivative there.

Chapter 4

4. Accumulate signed change and recover an amount

Goal: Use a definite integral for net change and split at sign changes when finding total distance.

For a continuous rate r(t), divide a time interval into small pieces. A rectangle with height r(t) and width Δt estimates the change on one piece. The definite integral is the limit of the sum as the largest piece width tends to zero. Negative rates contribute negative changes.

The Fundamental Theorem of Calculus connects this accumulation to derivatives. If r is continuous on [a, b] and F′ = r, then the integral from a to b of r equals F(b) − F(a). If A(x) is the integral from a to x of a continuous r(t), then A′(x) = r(x) at interior points.

A definite integral is a number with accumulated units, not a family of functions. An indefinite integral represents antiderivatives and includes an arbitrary constant. To recover a final amount from a rate, add the net change to the initial amount.

Worked example

The same motion has two different totals

An object has velocity v(t) = 2t − 4 m/s for 0 ≤ t ≤ 3 and initial position s(0) = 10 m. Find its displacement, final position, and total distance traveled.

  1. An antiderivative of 2t − 4 is t² − 4t. Evaluate at the upper endpoint minus the lower endpoint to find displacement.

    ∫03(2t−4) dt=[t2−4t]03=−3\int_0^3(2t-4)\,dt=[t^2-4t]_0^3=-3

  2. Add displacement to the initial position. A negative displacement means motion ended 3 meters in the negative direction from the starting point.

    s(3)=10+(−3)=7s(3)=10+(-3)=7

  3. Distance counts movement in either direction positively. Find where velocity changes sign: t = 2. It is negative before 2 and positive after 2.

    ∫03∣2t−4∣ dt=∫02(4−2t) dt+∫23(2t−4) dt\int_0^3|2t-4|\,dt=\int_0^2(4-2t)\,dt+\int_2^3(2t-4)\,dt

  4. The two parts are 4 meters and 1 meter. Add their magnitudes, rather than canceling them.

    4+1=5 m4+1=5\text{ m}

Displacement is −3 m, final position is 7 m, and total distance is 5 m. The magnitude of the displacement is only 3 m because backward and forward motion partially cancel.

Same motion, two ways to count area

Signed area → displacement

Signed area → displacementTwo graphs compare velocity and speed over 0 to 3 seconds. Velocity connects (0, −4), (2, 0), and (3, 2), in meters per second. The triangle below the axis has signed area −4 meters; the triangle above has area +1 meter. Their signed sum is −3 meters displacement. Speed connects (0, 4), (2, 0), and (3, 2). Both speed areas are positive, totaling 5 meters distance. Velocity changes sign at 2 seconds.−4−2240123Velocity (m/s)Time t (s)v(t) = 2t − 4−4 m+1 m

−4 m + 1 m = −3 m

Absolute area → distance

Absolute area → distanceTwo graphs compare velocity and speed over 0 to 3 seconds. Velocity connects (0, −4), (2, 0), and (3, 2), in meters per second. The triangle below the axis has signed area −4 meters; the triangle above has area +1 meter. Their signed sum is −3 meters displacement. Speed connects (0, 4), (2, 0), and (3, 2). Both speed areas are positive, totaling 5 meters distance. Velocity changes sign at 2 seconds.−4−2240123Speed (m/s)Time t (s)|v(t)| = |2t − 4|+4 m+1 m

4 m + 1 m = 5 m

For v(t) = 2t − 4 on 0 ≤ t ≤ 3 seconds, the striped region below the time axis contributes −4 m and the dotted region above contributes +1 m. Displacement is −3 m. Reflect the negative part upward to graph |v(t)|: distance is 4 + 1 = 5 m. Both plots use the same scale.

Avoid this trap: The integral of a function is signed area, not automatically geometric area. Similarly, |∫v(t)dt| generally differs from ∫|v(t)|dt when velocity changes sign. Taking an absolute value only at the end cannot recover canceled motion.

Practice on your own

Question 1

Evaluate the definite integral and check the result by the area of a trapezoid.

∫13(2x+1) dx\int_1^3(2x+1)\,dx
Answer, reasoning and error check for question 1

Answer: 10

Use F(x) = x² + x: F(3) − F(1) = 12 − 2 = 10. The graph has endpoint heights 3 and 7 and width 2, so the trapezoid area is (3 + 7) × 2/2 = 10.

Common mistake: 12 evaluates only the upper endpoint. 10 + C treats a definite integral as an indefinite one; the endpoint subtraction removes the arbitrary constant.

Need a review? Go to the matching chapter

Question 2

For this accumulation function, find A′(x) and A(2).

A(x)=∫0x(t2+1) dtA(x)=\int_0^x(t^2+1)\,dt
Answer, reasoning and error check for question 2

Answer: A′(x) = x² + 1; A(2) = 14/3

The integrand is continuous, so the Fundamental Theorem gives A′(x) = x² + 1. Separately, A(2) = [t³/3 + t] from 0 to 2 = 8/3 + 2 = 14/3.

Common mistake: x³/3 + x is A(x), not A′(x). The dummy variable t belongs inside the integral; the output derivative uses x.

Need a review? Go to the matching chapter

Before moving on: For any rate problem, write three labels before calculating: initial amount, net change, final amount. If the question asks for total distance or unsigned area, identify sign changes before integrating.

Mixed mastery check

Work these without viewing the examples. Include the domain or interval, units where relevant, and the reason your method applies. These test connections across chapters rather than speed.

Question 1

Find the limit and explain whether the original function is continuous at x = 4.

lim⁡x→4x2−16x−4\lim_{x\to4}\frac{x^2-16}{x-4}
Answer, reasoning and error check for question 1

Answer: The limit is 8. The original function is not continuous at 4 because it is not defined there.

For x ≠ 4, the quotient equals x + 4. Its limit is 8. Continuity also requires a defined value f(4) equal to that limit, which the original rule lacks.

Common mistake: “It is continuous because the limit exists” omits the function-value requirement.

Need a review? Go to the matching chapter

Question 2

For f(x) = (2x + 1)³, find f′(1) and the tangent line at x = 1. Use that line to estimate f(1.01).

Answer, reasoning and error check for question 2

Answer: f′(1) = 54; y − 27 = 54(x − 1); f(1.01) ≈ 27.54.

The chain rule gives f′(x) = 6(2x + 1)². At x = 1 the value is 27 and the slope is 54. A change of 0.01 adds approximately 54 × 0.01 = 0.54.

Common mistake: 27.27 misses the inner factor 2. Giving 27.54 as exact forgets that a tangent line is a local approximation to this curved graph.

Need a review? Go to the matching chapter

Question 3

An object has v(t) = 3t − 6 m/s on 0 ≤ t ≤ 4 and s(0) = 5 m. Find displacement, final position, and total distance. Explain why one can be zero.

Answer, reasoning and error check for question 3

Answer: Displacement 0 m; final position 5 m; total distance 12 m.

Integrating v gives [1.5t² − 6t] from 0 to 4 = 0. Velocity is negative until t = 2, then positive. The two intervals contribute −6 m and +6 m of displacement, so they cancel; their distances add to 12 m.

Common mistake: “Distance is zero because the integral is zero” counts cancellation as no movement. Total distance integrates speed, the absolute value of velocity.

Need a review? Go to the matching chapter

Move on when you can solve all three independently and explain the distinction in each: limit versus value, value versus derivative, and displacement versus distance. If a distinction is still unclear, revisit the linked chapter and solve a new version later. This is a learning checkpoint, not a certification of course or exam readiness.

Choose your next step

Repair algebra before adding more rules

If expansion, fractions, equations, or function inputs caused the mistakes, strengthen those first. Calculus rules become easier to use when the algebra no longer hides the concept.

Open the algebra learning unit

Continue through a complete first course

Next study product and quotient rules, trigonometric and exponential derivatives, derivative applications, and substitution for integrals. Follow a coherent textbook or course. Series belong to a later stage after the basic ideas of differentiation and integration are secure.

Compare calculus resources

Editorial note and corrections

The explanations and exercises in this unit are original OnlyMath learning material. Linked textbook sections provide optional further reading; their inclusion does not imply endorsement. English and Simplified Chinese cover the same questions and learning goals. Link checks were performed on October 8, 2026; external pages can change.

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