Lesson · Rates of Change and the Derivative · The derivative at a point

What the Quotients Approach

A column of exact quotients for halving steps gets closer to one number, and the gap to it shrinks with the step. That number is what the rate at the place means. The claim is about every step, not the rows on the page, and a row of the table need not equal the number it heads for. For a polynomial the claim is backed by the algebra of the previous lesson.

Open the interactive lesson Practice

halve h: 3, 5/2, 9/4, 17/8, 33/16 … the gap to 2 is 1, 1/2, 1/4, 1/8 … the quotients approach 2 no row equals 2; the claim covers every h t² at 1: quotient 2 + h, gap exactly h

What a column of quotients heads for

Core ideas

What this lesson establishes

Three ideas: the column, the gap, and what the word “approach” does and does not say.

The column of quotients

Fix a function and a place a. Start with a step h, then halve it again and again, and write the exact quotient for each. For t² at 1 the column is 3, 5/2, 9/4, 17/8, 33/16, 65/32, 129/64. The lab prints each as a fraction, so there is no rounding to blame for what the column does.

The gap shrinks

Subtract the number the column heads for from each row, and look at the size. Here the number is 2, and the gaps are 1, 1/2, 1/4, 1/8, …: each is exactly the step, because the quotient is 2 + h. As the step is halved the gap is halved, and it can be made smaller than any bound you name by halving enough times.

“Approach” is a claim about every step

The statement “the quotients approach 2” says that for every positive bound there is a step below which every quotient is within the bound of 2. A table of seven rows cannot say that. It shows seven cases and the pattern behind them; the claim is about all of them. Nor is any row equal to 2: the gap is h, never zero.

The material

A column of fractions, a number, and a claim

The table, the gap, the claim stated carefully, and the polynomial case where the claim is also algebra.

Take f(t) = t² at a = 1 and halve the step from h = 1 six times. The lab computes each quotient exactly.

f(t) = t², a = 1 h quotient gap to 2 1 3 1 1/2 5/2 1/2 1/4 9/4 1/4 1/8 17/8 1/8 1/16 33/16 1/16 1/32 65/32 1/32 1/64 129/64 1/64

The quotients fall, 3 down to 129/64, which is 2.015625 exactly. The gap column is the step column, exactly: the quotient is 2 + h, as the previous lesson found by algebra, so subtracting 2 leaves h.

Theorem · What the quotients approach

For f(t) = t² at a = 1, the exact quotients (f(1 + h) − f(1))/h approach 2 as h → 0: name any positive bound, and there is a step below which every quotient is within that bound of 2. When the quotients of a function at a place approach one number in this sense, that number is the rate of the function at the place. The lab's tile labels it f′(a), and the next lesson takes it as the definition of the derivative.

The word “approach” has a precise sense, and it is not “reach”. No row of the table equals 2. The last row is 129/64, and the next would be 257/128; there is always a row after it and always a gap. What the claim says is that the gap can be made as small as desired by taking the step small enough, and that is a statement about every step, not about the seven that happen to be printed.

Evidence and proof are different things

For this function the claim has a proof, and it is the algebra of the last lesson. The quotient is 2 + h for every nonzero h, so the gap to 2 is h in absolute value, which is less than any bound once h is. The table illustrates that; the algebra is what establishes it. For a function whose quotient does not simplify, the table is the evidence and the lesson states the claim, and later lessons say so where it applies.

Example · The cube at 1

For f(t) = t³ at a = 1 the quotient is 3 + 3h + h², from the expansion of (1 + h)³. At h = 1 it is 7, and at h = 1/64 it is 12481/4096.

The gap to 3 is 3h + h², which for h = 1/64 is 193/4096. Halving the step slightly more than halves the gap, because of the h² term, so the lab's gap column falls by a bit more than half each time. The column heads for 3.

Example · A flat place

For f(t) = t² at a = 0 the quotient is ((0 + h)² − 0)/h = h. The column is 1, 1/2, 1/4, …, and it heads for 0.

The rate at 0 is 0, which is what the graph of t² looks like there: the bottom of the bowl, neither rising nor falling. Here the gap to the limit is the quotient itself.

Written with an arrow, the claim for the first example is (f(1 + h) − f(1))/h → 2 as h → 0. The arrow reads as “goes to”, and it carries all the care of the paragraphs above. The next lesson gives the number its name at a point, and attaches a line to it.

Interactive

Difference quotients

Average rates as exact fractions

Halve the step and watch the gap

Each row halves the step. The last tile shows the number the column heads for, and the gap tile is the distance from the last row to it. Raise the number of halvings and the gap shrinks without ever reaching zero. The green line is the tangent.

Function and point

f(t)

first h

Halvings of h6

First quotient—

Last quotient—

Gap to f′(a)—

f′(a)—

Type a polynomial or a ratio of polynomials in t: t^2, t^3 - t, 1/t, (t+1)/(t-2).

Method

Reading a column of quotients

Four questions, asked of a table before it is trusted.

Write the quotient for each step

Halve the step each time and compute the exact quotient, as a fraction. Keep every row; a pattern can only be seen in a column.

Guess the number it heads for

Look at where the fractions are settling. For 3, 5/2, 9/4, 17/8 it is 2. If you cannot see it, simplify the quotient by algebra first.

Compute the gaps

Subtract the guess from each row. If the gaps shrink with the step, the guess is consistent with the table.

State the claim, and say what backs it

Say that the quotients approach the number as h → 0. Then say whether algebra backs the claim, as it does for a polynomial, or whether the table is the only evidence.

Approaching, not reaching

Score 0 / 0

Choose the answer you can justify.

The gaps for t² at 1

The quotient is 2 + h, so the gap to 2 is exactly the step. Seven rows, every one a fraction.

h = 1: 3 gap 1 h = 1/2: 5/2 gap 1/2 h = 1/4: 9/4 gap 1/4 h = 1/8: 17/8 gap 1/8 h = 1/16: 33/16 gap 1/16 h = 1/32: 65/32 gap 1/32 h = 1/64: 129/64 gap 1/64

Each gap is half the one above it, and none is zero. The table shows seven rows, and the claim is about every h, including the ones the table never reaches.

For a rehearsal, repeat this for t² at 2, where the quotient is 4 + h. Write the gaps to 4 before you look, then check them in the lab.

Common mistakes

Thinking the limit is the last row, or is reached at some stepThe last entry of a table is the quotient for the smallest step the table tried, not the number the column heads for. For t² at 1 the last row is 129/64, the number approached is 2, and the gap 1/64 is not zero. A smaller step gives a nearer row, and no step gives 2 itself.

Reading a pattern in a few rows as a proofSeven fractions heading for 2 are evidence. For t² the algebra turns the evidence into a proof, because the quotient is exactly 2 + h. For a function with no such simplification, the table remains evidence, and a lesson that relies on it should say so.

Expecting the gap to halve every timeThe gap halves for t² at 1 because the quotient is 2 + h. For t³ at 1 the gap is 3h + h², which falls to slightly less than half each time. How the gap shrinks depends on the function; that it shrinks is the claim.

Check your understanding

✓

Finish when you can produce the column of exact quotients for halving steps, state the number they approach, and say in one sentence what that claims and what the table shows.

You should be able to write the exact quotient for a polynomial at a place, list the first several rows, compute the gaps, and say that the claim covers every step while the table covers only the rows shown.

Note: The number the quotients approach at one place gets its definition next, and it comes with a line: “The Derivative at a Point and the Tangent Line”.