AP Calculus AB and BC

Secant to Tangent: The Limit Definition, Made Physical

Q: Why can't I just plug in h = 0 to find the derivative?

Substituting h = 0 turns the difference quotient into \frac{f(a) - f(a)}{0} = \frac{0}{0}, which is undefined, so the expression genuinely has no value at h = 0. The limit as h \to 0 sidesteps this by asking what value the quotient approaches for h near 0 but not equal to 0.

Q: Are the two limit definitions of the derivative the same?

Yes. \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} and \lim_{x \to a} \frac{f(x) - f(a)}{x - a} are equal whenever the derivative exists. Substitute x = a + h: as x \to a, then h \to 0, and the two expressions match term for term. Use whichever form makes a given problem easier.

The derivative f'(a) is the slope of the tangent line at x=a. You get it from the slope of a secant line through two nearby points: as h approaches 0, the difference quotient (f(a+h)-f(a))/h stops being an average rate of change and becomes the instantaneous rate at the point.

f(x) =

secant (slope = difference quotient)tangent at a (slope = f'(a))

h1.00

\frac{f(1) - f(0)}{1} = \frac{1 - 0}{1} = 1

1.00

difference quotient

1.0000

f'(a) exact

0.0000

Drag the amber point (or shrink h) and watch the secant pivot into the ghost tangent.

A single point on a graph has no rate of change. Speed, slope, and rate all compare two things, so at one isolated instant there is nothing to compare. The limit definition of the derivative works around this: start with two points, measure the average rate of change between them, then slide the second point toward the first until the gap vanishes. The number that the average rate settles on is $f'(a)$ , the instantaneous rate of change at $x = a$ (Topics 2.1 and 2.2).

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}

Geometrically, $\frac{f(a+h) - f(a)}{h}$ is the slope of the secant line through the two points $(a, f(a))$ and $(a+h, f(a+h))$ . Shrinking $h$ walks the second point along the curve toward the first, and the secant line pivots. In the limit the two points merge and the secant becomes the tangent line at $x = a$ . Average rate of change over the interval between $a$ and $a+h$ turns into instantaneous rate of change at the single point $a$ .

Method check: is this limit secretly a derivative?

The most tested payoff of this definition is running it in reverse. If you meet a limit like $\lim_{h \to 0} \frac{(2+h)^3 - 8}{h}$ and it looks like a dead end, check whether it matches the difference quotient. Here $(2+h)^3$ matches $f(a+h)$ with $f(x) = x^3$ and $a = 2$ . Check the constant: $f(a) = 2^3 = 8$ matches the $-8$ in the numerator, so the whole limit is just $f'(2) = 3(2)^2 = 12$ . Recognizing the form lets you skip the algebra and differentiate instead (Topic 2.7, LIM-3.A.1).

Notice you can never simply set $h = 0$ . Substituting gives $\frac{f(a) - f(a)}{0} = \frac{0}{0}$ , which is undefined: the difference quotient does not exist exactly at the point. That indeterminate form is the whole reason calculus needs limits. The limit lets the ratio have a value that the expression itself never reaches (Topic 1.1).

How to read it

The screen shows a curve $y = f(x)$ with a fixed anchor point at $x = a$ and a second point at $x = a + h$ that you control. A straight secant line runs through both points, and a readout reports the current value of $h$ and the current difference quotient (the secant slope).

Drag $h$ toward 0. Watch the movable point slide down the curve toward the anchor and the secant line pivot as it goes.
Keep your eye on the difference-quotient readout. As $h$ shrinks, the number stops wandering and homes in on one value: that value is $f'(a)$ , the slope of the tangent.
Notice the secant never needs $h$ to actually hit 0. It gets arbitrarily close, which is exactly what $\lim_{h \to 0}$ means.

Now drag $h$ to a negative value so the second point sits to the left of the anchor. The secant approaches the same line and the readout approaches the same number. Left and right give one shared limit, which is what it means for $f'(a)$ to exist. Where a graph has a corner, the two sides would disagree and no single tangent slope would exist (Topic 2.4).

What the readout is proving

The label on the secant slope is average rate of change over the interval between $a$ and $a+h$ . As you close the gap, that same number becomes the instantaneous rate of change at $a$ . You are watching one quantity change its meaning in real time, which is the entire content of the limit definition.

Frequently asked questions

What is the difference between a secant line and a tangent line?

A secant line is the line through two chosen points on a curve, and its slope is the average rate of change between them. A tangent line meets the curve at the point of tangency, matching the curve's direction there (it may still cross the curve elsewhere), and its slope is the instantaneous rate of change at that point, which equals $f'(a)$ . The tangent is the limiting position of the secant as the two points slide together.

Why can't I just plug in h = 0 to find the derivative?

Substituting $h = 0$ turns the difference quotient into $\frac{f(a) - f(a)}{0} = \frac{0}{0}$ , which is undefined, so the expression genuinely has no value at $h = 0$ . The limit as $h \to 0$ sidesteps this by asking what value the quotient approaches for $h$ near 0 but not equal to 0.

Are the two limit definitions of the derivative the same?

Yes. $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ and $\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ are equal whenever the derivative exists. Substitute $x = a + h$ : as $x \to a$ , then $h \to 0$ , and the two expressions match term for term. Use whichever form makes a given problem easier.