Single-Variable-Calculus · Unit 1 · Video 2 · Interactive Practice

Secant Lines and the Limit Definition of the Derivative

IKey Formulas

Formula	Name	What it says
$Q = \big(x_0 + \Delta x,\ f(x_0 + \Delta x)\big)$	The moving point	$P$ stays fixed; $\Delta x$ alone locates $Q$
$\Delta f = f(x_0 + \Delta x) - f(x_0)$	Rise from $P$ to $Q$	Height of $Q$ minus height of $P$
$m_{PQ} = \dfrac{\Delta f}{\Delta x}$	Slope of the secant $PQ$	Rise over run across a finite step
$f'(x_0) = \lim\limits_{\Delta x \to 0} \dfrac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$	Limit definition of the derivative	Slope of the tangent line at $x_0$

Key Insight: The tangent line is the limit of the secant lines $PQ$ as $Q \to P$ . Because $Q$ sits at $x_0 + \Delta x$ , saying $Q \to P$ is exactly saying $\Delta x \to 0$ — geometry on the left, a computable limit on the right.

IIVisualization 1 — The Secant Rotates Into the Tangent

As $Q$ slides toward $P$ on $f(x) = \tfrac{1}{4}x^2$ , the secant slope settles on one number: $f'(2)$ .

Δx 2.000

IIIVisualization 2 — Why "Meets the Graph Once" Fails

A genuine tangent line may cut back through its own curve elsewhere; tangency is a local condition at $P$ .

IVVisualization 3 — Building the Formula, Step by Step

Each step attaches one symbol to the picture, ending at the limit that defines $f'(x_0)$ .

Step 1

Step 1 — Name the two points

P = \big(x_0,\ f(x_0)\big) = (2,\ 1)

Q = \big(x_0 + \Delta x,\ f(x_0 + \Delta x)\big) = \big(3,\ \tfrac{9}{4}\big)

Step 2 — The run

\Delta x = (x_0 + \Delta x) - x_0 = 3 - 2 = 1

Step 3 — The rise

\Delta f = f(x_0 + \Delta x) - f(x_0) = \tfrac{9}{4} - 1 = \tfrac{5}{4}

Step 4 — Slope of the secant $PQ$

\frac{\Delta f}{\Delta x} = \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} = \frac{5/4}{1} = \tfrac{5}{4}

For any

\Delta x

\frac{\tfrac{1}{4}(2 + \Delta x)^2 - 1}{\Delta x} = \frac{\Delta x + \tfrac{1}{4}(\Delta x)^2}{\Delta x} = 1 + \frac{\Delta x}{4}

Step 5 — Let $\Delta x \to 0$

f'(2) = \lim_{\Delta x \to 0} \frac{\Delta f}{\Delta x} = \lim_{\Delta x \to 0} \left(1 + \frac{\Delta x}{4}\right) = 1

💡 At $\Delta x = 0$ the quotient reads $\tfrac{0}{0}$ , so the limit only exists once the algebra cancels a factor of $\Delta x$ — here $\Delta x + \tfrac14(\Delta x)^2$ over $\Delta x$ . Next: the same cancellation for $f(x) = 1/x$ .

VQuiz Questions

Problem 1 · Slope of a Secant Line

Given: $f(x) = x^2$ , with $P = \big(1, f(1)\big)$ and $Q = \big(1.5, f(1.5)\big)$ — find the slope of the secant line $PQ$ .

✅ Correct!

\Delta f = 2.25 - 1 = 1.25

and

\Delta x = 0.5

, so the secant slope is

1.25 / 0.5 = 2.5

❌ That is $\Delta f$ , not the slope.

1.25

is the rise alone; slope is rise over run, so divide by

\Delta x = 0.5

❌ That is the tangent slope, not the secant slope.

f'(1) = 2

is the limiting value; across a finite step

\Delta x = 0.5

the secant is steeper.

❌ Not quite.

3

is the tangent slope at the other endpoint,

x = 1.5

. Compute

\Delta f = f(1.5) - f(1)

first, then divide by

\Delta x

Show solution

Here $x_0 = 1$ and $\Delta x = 0.5$ .

\Delta f = f(x_0 + \Delta x) - f(x_0) = (1.5)^2 - (1)^2 = 2.25 - 1 = 1.25

\frac{\Delta f}{\Delta x} = \frac{1.25}{0.5} = 2.5

Note the secant slope $2.5$ sits between the tangent slopes at the two endpoints, $f'(1) = 2$ and $f'(1.5) = 3$ — a finite step averages the steepness along the way.

Problem 2 · What Makes a Line Tangent

Which statement correctly characterizes the tangent line to the graph of $f$ at the point $P$ ?

✅ Correct!

P

is held fixed,

Q

is the variable point, and the tangent is what the secants

PQ

approach as

Q \to P

❌ Counting intersections is not the test. On a wiggly curve the true tangent at

P

can cut back through the graph several more times — as Visualization 2 shows.

❌ A tangent may cross the curve. At an inflection point the tangent passes straight through the graph —

f

has three of them inside the drag range of Visualization 2 — so "never crosses" is not what tangency means.

❌ That is a secant, not the tangent. Any fixed

\Delta x \neq 0

still gives a line through two distinct points; the tangent needs the limit

\Delta x \to 0

Show solution

A secant line joins two points of the graph, $P$ and $Q$ . Holding $P$ fixed and letting $Q$ slide toward $P$ , the secant rotates; the line it rotates onto is the tangent:

\text{tangent at } P = \lim_{Q \to P} (\text{secant } PQ)

Why the other descriptions fail:

"Meets the graph once" and "never crosses": a tangent line at a point of an oscillating curve can meet the graph at many other points, and a tangent at an inflection point crosses the curve right at $P$ .
"Slope $\Delta f / \Delta x$ for a small fixed $\Delta x$ ": that is a secant slope. It approximates $f'(x_0)$ , but the derivative is defined by the limit, not by any one small step.

Problem 3 · Run the Definition on $f(x) = 3x^2$

Given: $f(x) = 3x^2$ at $x_0 = 2$ — simplify the difference quotient $\dfrac{f(2 + \Delta x) - f(2)}{\Delta x}$ , then take the limit.

Simplified difference quotient

Value of $f'(2)$

✅ Correct!

\Delta f = 12\,\Delta x + 3(\Delta x)^2

, so the quotient is

12 + 3\Delta x

, and letting

\Delta x \to 0

leaves

f'(2) = 12

❌ That is $\Delta f$ , not $\Delta f / \Delta x$ . You still owe one division by

\Delta x

— and it divides both terms.

❌ Check the expansion.

3(2 + \Delta x)^2 = 12 + 12\,\Delta x + 3(\Delta x)^2

; subtract

f(2) = 12

, then divide every remaining term by

\Delta x

❌ Check the limit. With the quotient simplified to

12 + 3\Delta x

, sending

\Delta x \to 0

kills only the

3\Delta x

term.

❌ $\Delta f \to 0$ , but the ratio does not. Both

\Delta f

and

\Delta x

shrink to zero together; their ratio settles on a nonzero number.

❌ A derivative is a number, not an expression in $\Delta x$ . The limit has to be taken — substitute

\Delta x = 0

into the simplified quotient.

Show solution

Step 1 — Write $\Delta f$ explicitly:

\Delta f = 3(2 + \Delta x)^2 - 3(2)^2 = 3\big(4 + 4\Delta x + (\Delta x)^2\big) - 12

\Delta f = 12 + 12\,\Delta x + 3(\Delta x)^2 - 12 = 12\,\Delta x + 3(\Delta x)^2

Step 2 — Divide by $\Delta x$ (legal because $\Delta x \neq 0$ in the limit process):

\frac{\Delta f}{\Delta x} = \frac{12\,\Delta x + 3(\Delta x)^2}{\Delta x} = 12 + 3\Delta x

Step 3 — Take the limit:

f'(2) = \lim_{\Delta x \to 0} \big(12 + 3\Delta x\big) = 12

Cancelling the $\Delta x$ is the whole trick: before it, the quotient is $\tfrac{0}{0}$ at $\Delta x = 0$ ; after it, the limit is just a substitution.

Problem 4 · Reading a Derivative Off the Secants

Given: secant slopes of a function $g$ measured from $x_0 = 1$ : at $\Delta x = 0.1$ the slope is $6.31$ ; at $\Delta x = 0.01$ it is $6.0301$ ; at $\Delta x = 0.001$ it is $6.003001$ . What is $g'(1)$ ?

✅ Correct! The secant slopes are

6 + 3\Delta x + (\Delta x)^2

, and the value they close in on as

\Delta x \to 0

6

❌ That is a secant slope, not the limit. Every entry in the list belongs to a finite

\Delta x

; the derivative is the number the entries approach as

\Delta x \to 0

❌ Not zero.

\Delta f

shrinks toward

0

, but the list shows the ratio

\Delta f / \Delta x

holding steady near

6

❌ Not quite. Track the excess over a round number:

0.31

, then

0.0301

, then

0.003001

— the excess is dying out.

Show solution

Line up the measurements against $\Delta x$ :

\Delta x = 0.1 \Rightarrow 6.31, \qquad \Delta x = 0.01 \Rightarrow 6.0301, \qquad \Delta x = 0.001 \Rightarrow 6.003001

Each slope is $6$ plus a remainder that shrinks by roughly a factor of $10$ every time $\Delta x$ does — in fact the quotient here is exactly

\frac{\Delta g}{\Delta x} = 6 + 3\Delta x + (\Delta x)^2,

so $6.31 = 6 + 0.3 + 0.01$ and $6.0301 = 6 + 0.03 + 0.0001$ . Taking $\Delta x \to 0$ removes every $\Delta x$ term:

g'(1) = \lim_{\Delta x \to 0}\big(6 + 3\Delta x + (\Delta x)^2\big) = 6

The derivative is the value the secant slopes converge to, never one of the measured slopes themselves.

Solved: 0 / 4