How does a limit define an instantaneous rate?
You will be able to: Evaluate a derivative at a point from either limit definition.
BC foundation: Unit 2 shares its derivative foundations with AB. Connect each rule to rates, graphs and its domain conditions. Use the written challenges to explain your reasoning. Chain-rule compositions, implicit and inverse differentiation, and advanced applications come later.
How does a limit define an instantaneous rate?
The vehicle’s average rates near 2 seconds are 4.1 m/s on the interval ending at 2.1 and 3.9 m/s on the interval beginning at 1.9. Both point toward 4 m/s at the shared time.
A useful starting point: How does an interval become a rate? →
Words and symbols before equations
- Derivative at a
- The finite limit of difference quotients at input a, when it exists.
- Instantaneous rate
- Limiting rate at one input, not an average across a fixed interval.
- Two-sided limit
- A common value approached from smaller and larger inputs.
- f′(a)
- Read f prime of a; a number representing local rate.
What this picture assumes
Original equation-driven model; readouts are rounded. Graphs have labeled linear scales and finite sampled windows; algebra supplies exact conclusions. Same position rule s=t²+1. The limit of the secant rate is 4 m/s; the green tangent depicts that exact limiting rate.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- h=0.1 s; rise=0.41 m; run=0.1 s; secant rate=4.1 m/s. Exact derivative at 2 s=4 m/s. Blue: position; orange: secant; green: exact tangent. Position s(2)=5 m is a different quantity.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
For s(t)=t²+1, s′(2)=lim h→0 [s(2+h)−s(2)]/h. Algebra gives 4+h whenever h≠0, so the limit is 4.
The equivalent form lim t→2 [s(t)−s(2)]/(t−2) uses t instead of h. Since h=t−2, both definitions inspect exactly the same nearby intervals.
The limiting number has rate units m/s. It is not s(2)=5 m and does not require a zero-length quotient.
At an interior point the finite left and right limits must agree. A one-sided estimate alone cannot establish that a derivative exists.
A worked example, step by step
Use the definition to find f′(3) for f(x)=x²+1.
- Write lim h→0 {[(3+h)²+1]−10}/h.
- Expand the numerator to 6h+h².
- For h≠0 simplify to 6+h.
- The two-sided limit is 6, so f′(3)=6.
A derivative is a limit of quotients; substituting h=0 into the original quotient is undefined.
Is an extremely small h the same as the limit?
Compare with an explanation
No. It produces one nearby quotient. The expression 4+h establishes the exact limit as h approaches zero.
Predict. Change one thing. Explain.
Reduce the nonzero interval from each side. Explain why every displayed rate differs from 4 although their exact limit is 4.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
h=0.1 s; rise=0.41 m; run=0.1 s; secant rate=4.1 m/s. Exact derivative at 2 s=4 m/s. Blue: position; orange: secant; green: exact tangent. Position s(2)=5 m is a different quantity.
Original equation-driven model; readouts are rounded. Graphs have labeled linear scales and finite sampled windows; algebra supplies exact conclusions. Same position rule s=t²+1. The limit of the secant rate is 4 m/s; the green tangent depicts that exact limiting rate.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using a difference quotient, tangent slope, or derivative rule with its domain conditions. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionFind the derivative of g(x)=3x+2 at a=4 using the limit definition and explain why no division by zero is performed.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: g(4)=14.
- 1 point: g(4+h)−g(4)=3h.
- 1 point: The nonzero-h quotient is 3.
- 1 point: Its limit is 3; zero is approached but never substituted into the original quotient.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1When does the ordinary derivative exist?
When the difference quotient has a finite two-sided limit at an interior point.
RECALL 2How are x and h related?
x=a+h, so h=x−a.
RECALL 3Is f′(a) a height?
No. It is a local rate or tangent slope.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How does a limit define an instantaneous rate?
- f′(a)=lim h→0 [f(a+h)−f(a)]/h.
- Also f′(a)=lim x→a [f(x)−f(a)]/(x−a), if the finite limit exists.
Remember: A derivative is a limit of quotients; substituting h=0 into the original quotient is undefined.
Conditions: Original equation-driven model; readouts are rounded. Graphs have labeled linear scales and finite sampled windows; algebra supplies exact conclusions. Same position rule s=t²+1. The limit of the secant rate is 4 m/s; the green tangent depicts that exact limiting rate.
Refresh Kid · AP Calculus BC Unit 2 · Objectives CHA-2.A, CHA-2.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.1, CHA-2.A, CHA-2.B. CED effective Fall 2020, with Fall 2026 clarifications, checked September 17, 2026. Unit 2 has 10 official topics. Focused lesson titles and questions are original Refresh Kid teaching material.
Derivative definitions require finite two-sided limits at interior inputs. Graphs and finite tables supply evidence and estimates, not universal proofs. Continuity is necessary but insufficient for differentiability. Rules retain their original domain restrictions, and trigonometric inputs use radians. Chain-rule compositions, implicit differentiation, inverse-function differentiation and L’Hôpital’s rule are deferred to later units. Piecewise joins are checked for continuity before their side slopes are compared.
The Organic Chemistry Tutor video titles, creators and descriptions were checked; full videos were not reviewed. Khan Academy’s unit destination was checked; its JavaScript lesson content was not fully readable by the research tool. OpenStax Sections 3.1, 3.3 and 3.5 were consulted for conceptual cross-checking. No creator scripts, questions, diagrams or artwork were copied. Refresh Kid is not endorsed by these providers.
GitHub’s 3D website collection and its camera-control example informed optional spatial inspection. Graph geometry is generated from the stated original equations; self-hosted Three.js retains its MIT license. The optional 3D model uses separate planes for f and f′. Depth distinguishes panels, not an additional variable; camera rotation does not change their values. Use the complete labeled 2D graphs and text to read precise coordinates and slopes.
Independent teacher review and observation of students remain pending. Checks do not certify scientific accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus BC questions and scoring guides. The archive spans multiple units; no entire exam question is assigned to this lesson.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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