A. Rate of Change · Limits and Rates of Change

The Difference Quotient as a Limit

A1.5About 70 minutes

Learning goals

I can write the average rate of change over [a, a + h] as the difference quotient (f(a + h) − f(a))/h.
I can write the instantaneous rate of change at x = a as lim (h → 0) (f(a + h) − f(a))/h.
I can recognize which function, point and graph feature a given difference-quotient limit describes.
I can evaluate difference-quotient limits and use them to write tangent equations.

Words to know

difference quotient
(f(a + h) − f(a))/h, the slope of the secant from x = a to x = a + h, with h ≠ 0.

h
The width (run) of the interval. h may be positive ([a, a + h]) or negative ([a + h, a]).

instantaneous rate of change at a
lim (h → 0) (f(a + h) − f(a))/h, when this limit exists.

slope of the tangent
The same limit, seen on the graph: the slope of the tangent line at (a, f(a)).

tangent line equation
y − f(a) = m(x − a), where m is the slope of the tangent at x = a.

Putting the pieces together

You now have both halves of the central idea of calculus:

From the first lessons: the average rate of change over an interval is the slope of a secant, and shrinking the interval gives the instantaneous rate (the slope of the tangent).
From the limit lessons: limits describe what a quantity approaches, and algebra can evaluate them exactly, even when substitution gives 0/0.

This lesson writes the first idea in the language of the second.

The difference quotient

Take a function that is smooth on the interval from a to a + h. The two endpoints of the graph are P(a, f(a)) and Q(a + h, f(a + h)). The run is (a + h) − a = h and the rise is f(a + h) − f(a). So

average rate of change on [a, a + h] = (f(a + h) − f(a))/h

This expression is called the difference quotient (a quotient of differences). Now let the interval shrink: h → 0, from either side. The secant turns into the tangent, and

instantaneous rate of change at x = a = lim (h → 0) (f(a + h) − f(a))/h

An equivalent form uses x in place of a + h: lim (x → a) (f(x) − f(a))/(x − a). Both say "the limit of secant slopes as the second point approaches the first".

When h is negative, the interval is [a + h, a], which lies to the left of a. The same formula still gives that secant's slope.

The difference quotient is rise over run for the secant PQ: the run is h and the rise is f(a + h) − f(a). Its limit as h → 0 is the slope of the tangent at P.

Worked example

Example 1: Average and instantaneous rates from one expression

For f(x) = x² − 3x, simplify the difference quotient at a = 2. Use it to find the average rate on [2, 2.5] and the instantaneous rate at x = 2.

f(2) = 4 − 6 = −2.
f(2 + h) = (2 + h)² − 3(2 + h) = 4 + 4h + h² − 6 − 3h = −2 + h + h².
f(2 + h) − f(2) = h + h², so the difference quotient is (h + h²)/h = 1 + h (h ≠ 0).
Average rate on [2, 2.5]: h = 0.5 gives 1 + 0.5 = 1.5. (Check: f(2.5) = 6.25 − 7.5 = −1.25, and (−1.25 − (−2))/0.5 = 1.5 ✓.)
Instantaneous rate: lim (h → 0) (1 + h) = 1.

Average rate on [2, 2.5]: 1.5. Instantaneous rate at x = 2: 1. One simplified expression, 1 + h, gives every average rate starting at 2, and its limit gives the instantaneous rate.

Worked example

Example 2: What does the limit tell you? (based on a curriculum sample)

What does lim (h → 0) ((4 + h)² − 16)/h indicate about the graph of f(x) = x²?

Match the pattern (f(a + h) − f(a))/h: f(4 + h) = (4 + h)² and f(4) = 16, so a = 4.
Evaluate: ((16 + 8h + h²) − 16)/h = (8h + h²)/h = 8 + h → 8.
So the limit is the slope of the tangent to y = x² at the point (4, 16).
Tangent: y − 16 = 8(x − 4), so y = 8x − 16.

The limit equals 8: the graph of y = x² has a tangent of slope 8 at (4, 16), the line y = 8x − 16. The function's instantaneous rate of change at x = 4 is 8.

Secants from (4, 16) on y = x² with h = 2 and h = 1, and the tangent they approach. The secant slope is 8 + h, which approaches 8.

Worked example

Example 3: Reading a limit backwards

Each limit is an instantaneous rate of change. Identify f and a, and evaluate.

(a) lim (h → 0) (√(9 + h) − 3)/h (b) lim (x → 3) (x³ − 27)/(x − 3)

(a) f(9 + h) = √(9 + h) and f(9) = 3 = √9, so f(x) = √x and a = 9. Rationalize: h/(h(√(9 + h) + 3)) = 1/(√(9 + h) + 3) → 1/6.
(b) This is the form (f(x) − f(a))/(x − a) with f(x) = x³ and a = 3 (since 3³ = 27). Factor: x² + 3x + 9 → 27.

(a) The slope of y = √x at (9, 3) is 1/6. (b) The slope of y = x³ at (3, 27) is 27. (Other answers are possible: for (a), f(x) = √(9 + x) with a = 0 also works.)

Worked example

Example 4: Velocity as a limit

A stone is dropped from a bridge. It falls s(t) = 4.9t² metres in t seconds. Find its velocity at t = 3 s.

Velocity is the instantaneous rate of change of distance: v(3) = lim (h → 0) (s(3 + h) − s(3))/h.
s(3 + h) = 4.9(9 + 6h + h²) = 44.1 + 29.4h + 4.9h², and s(3) = 44.1.
Difference quotient: (29.4h + 4.9h²)/h = 29.4 + 4.9h.
Limit as h → 0: 29.4.

The stone is falling at 29.4 m/s at t = 3 s. (The average velocity over [3, 3.5] would be 29.4 + 4.9(0.5) = 31.85 m/s.)

Read a limit like a sentence

"lim (h → 0) (f(a + h) − f(a))/h = m" says: the graph of f has a tangent at (a, f(a)) with slope m, and f is changing at m units of output per unit of input at x = a. Always name the function, the point and the meaning.

Just for fun

Riddle: what gets smaller every time you look at it, never quite disappears, and is only allowed to leave at the very end?

The h in a difference quotient.

Hands-on activity

Rolling down a ramp

You need: a long board or a sturdy shelf about 1 to 2 m long, a few books to prop one end up about 10 cm, a toy car or a ball, a measuring tape, masking tape, and a phone that records video.

Mark the ramp with tape every 20 cm from the top.
Film the car rolling from rest at the top, with the marks visible. Use slow motion if your phone has it.
Play the video frame by frame. Record the time when the car passes each mark (most phones show time to 0.01 s or let you count frames).
Make a table of distance d against time t. Find the average velocity over several intervals that start at the same time t = a, with shorter and shorter widths h.
Fit a model d = kt² by choosing k from one data point, and use the difference quotient of your model to find the instantaneous velocity at t = a.

What to look for: distances grow faster and faster (the car speeds up). Average velocities over [a, a + h] get smaller as h shrinks and settle near one value. For a model d = kt², the difference quotient simplifies to 2ka + kh, so the instantaneous velocity is 2ka. Your estimate from the data should be close to that.

Your turn

For f(x) = x², write and simplify the average rate of change on [3, 3 + h].

Use your answer to find the average rate on [3, 3.1].

Find the instantaneous rate of change of f(x) = x² at x = 3.

What does lim (h → 0) ((1 + h)³ − 1)/h represent? Evaluate it.

Identify f and a, and evaluate: lim (h → 0) (√(25 + h) − 5)/h.

What does lim (x → 2) (x² − 4)/(x − 2) represent? Evaluate it.

Write the equation of the tangent to y = x² at x = 3.

More practice: warm-up

Write the interval [a, a + h] for a = 5 and h = 0.2.

For a = 5 and h = −0.1, which interval does the difference quotient describe?

Write the difference quotient for a function f at a = 1.

A difference quotient simplifies to 10 + h. What is the instantaneous rate of change?

Distance s is in metres and time t in seconds. What are the units of lim (h → 0) (s(t + h) − s(t))/h?

More practice: on your own

For f(x) = 2x² + 1, find the instantaneous rate of change at x = −1 using the difference quotient.

Identify f and a, and evaluate: lim (h → 0) [1/(2 + h) − 1/2]/h.

The limit lim (h → 0) (2^(3 + h) − 8)/h is an instantaneous rate. Identify f and a, and estimate it with h = 0.001.

For the dropped stone s(t) = 4.9t², find the velocity at t = 2 s, and the average velocity on [2, 2.5].

A spherical balloon has volume V(r) = (4/3)πr³. Write the rate of change of volume at r = 10 cm as a limit and evaluate it.

Challenge

Identify f and a in lim (h → 0) (sin(π/2 + h) − 1)/h. What should the value be, judging by the graph? Check with h = 0.01.

lim (h → 0) ((a + h)² − a²)/h = 10. Find a and the point on y = x² that this describes.

Show that lim (h → 0) ((a + h)³ − a³)/h = 3a². Then find the points on y = x³ where the tangent is parallel to the line y = 12x.

Check your understanding

Quiz

The expression (f(a + h) − f(a))/h, with h ≠ 0, represents:

What does lim (h → 0) ((4 + h)² − 16)/h = 8 tell you?

lim (h → 0) (√(16 + h) − 4)/h is the instantaneous rate of change of which function, at which point?

For a function, (f(2 + h) − f(2))/h simplifies to 7 − 3h. What is the average rate of change on [2, 3]?

For s(t) = 4.9t², the difference quotient at t = 3 is 29.4 + 4.9h. What is the velocity at t = 3?

Which limit is equivalent to lim (h → 0) (f(a + h) − f(a))/h?

For parents and teachers

Big idea (A1.5): for a function smooth on [a, a + h], the difference quotient (f(a + h) − f(a))/h is the average rate of change on that interval, and its limit as h → 0 is the instantaneous rate of change at a (the slope of the tangent).

How to help: Write a limit such as lim (h → 0) ((5 + h)² − 25)/h on paper and ask your teen three questions: Which function? Which point? What does the answer tell you about the graph? (y = x², the point (5, 25), and a tangent slope of 10.)

Common difficulties:

Treating the difference quotient as f(a + h) − f(a)/h (missing brackets).
Writing f(a + h) as f(a) + h. For f(x) = x², f(a + h) = (a + h)², not a² + h.
Substituting h = 0 before simplifying.
Not stating what the limit means: a slope, a rate, and its units.