Math Core

You know how to find the slope of a line: rise over run. But what is the slope of a curve at a single point, like the steepness of a parabola at $x = 1$ , or a car's speed at one exact instant? Answering that question is where calculus begins, and limits are exactly the tool it needs.

Why one point isn't enough

To compute a slope you need two points. A tangent line at a point $P$ on a curve touches the curve at $P$ and points in the same direction as the curve there. But you only know one point on it, $P$ itself, so rise over run gives $\dfrac{0}{0}$ .

The fix is to approximate. Pick a second point $Q$ on the curve, close to $P$ , and draw the line through both. A line through two points of a curve is called a secant line, and you can compute its slope. Then slide $Q$ toward $P$ and watch what happens to the secant slopes.

Secant slopes close in on the tangent slope

Take $f(x) = x^2$ and the point $P = (1, 1)$ . Let the second point be $Q = \left(1 + h, (1 + h)^2\right)$ , where $h$ is the horizontal distance from $P$ to $Q$ . The slope of the secant line is

m_{\text{sec}} = \frac{f(1 + h) - f(1)}{(1 + h) - 1} = \frac{(1 + h)^2 - 1}{h}.

Here are the secant slopes for several values of $h$ , positive and negative.

$h$	$1$	$0.5$	$0.1$	$0.01$	$-0.01$	$-0.1$	$-0.5$
$m_{\text{sec}}$	$3$	$2.5$	$2.1$	$2.01$	$1.99$	$1.9$	$1.5$

As $h$ shrinks toward $0$ from either side, the secant slopes approach $2$ . So the slope of the tangent line at $(1, 1)$ is $2$ , and the tangent line is $y - 1 = 2(x - 1)$ , or $y = 2x - 1$ .

The parabola y = x², the secant through P(1, 1) and Q(2, 4) with slope 3 (dashed), and the tangent at P with slope 2.Open in grapher →

Algebra confirms the table exactly. Expand and simplify before letting $h \to 0$ :

\frac{(1 + h)^2 - 1}{h} = \frac{1 + 2h + h^2 - 1}{h} = \frac{h(2 + h)}{h} = 2 + h \longrightarrow 2.

The definition

Definition

Slope of the tangent line

The slope of the tangent line to the graph of $f$ at the point $(a, f(a))$ is

m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h},

provided this limit exists. The fraction inside is called the difference quotient.

This number is also called the slope of the curve at $x = a$ . An equivalent form uses a second point $(x, f(x))$ and lets $x \to a$ :

m = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}.

Every one of these limits starts as $\dfrac{0}{0}$ when you substitute, because the rise and the run both shrink to $0$ . That's why the algebra tools from the limit laws (expand, factor, rationalize, combine fractions) are exactly what you need here.

Finding a tangent line

Compute $f(a)$ and $f(a + h)$ .
Simplify the difference quotient $\dfrac{f(a + h) - f(a)}{h}$ until the $h$ in the denominator cancels.
Let $h \to 0$ to get the slope $m$ .
Write the line in point-slope form: $y - f(a) = m(x - a)$ .

Common mistake

Don't set $h = 0$ at the start. The difference quotient is $\dfrac{0}{0}$ at $h = 0$ . Simplify first, cancel the factor of $h$ , and only then let $h$ approach $0$ .

Worked example: A tangent line to a parabola

Find the slope of the tangent line to $f(x) = x^2 - 3x$ at $x = 2$ , and write its equation.

Solution. $f(2) = 4 - 6 = -2$ . Next,

f(2 + h) = (2 + h)^2 - 3(2 + h) = 4 + 4h + h^2 - 6 - 3h = -2 + h + h^2.

The difference quotient is

\frac{(-2 + h + h^2) - (-2)}{h} = \frac{h + h^2}{h} = 1 + h,

which approaches $1$ . So $m = 1$ , and the tangent line is $y - (-2) = 1(x - 2)$ , or $y = x - 4$ .

Worked example: A tangent line to a hyperbola

Find the equation of the tangent line to $f(x) = \dfrac{1}{x}$ at $x = 2$ .

Solution. $f(2) = \dfrac{1}{2}$ . Combine fractions in the numerator of the difference quotient:

\frac{\frac{1}{2 + h} - \frac{1}{2}}{h} = \frac{\frac{2 - (2 + h)}{2(2 + h)}}{h} = \frac{-h}{2h(2 + h)} = \frac{-1}{2(2 + h)}.

As $h \to 0$ this approaches $-\dfrac{1}{4}$ . The tangent line is $y - \dfrac{1}{2} = -\dfrac{1}{4}(x - 2)$ , which simplifies to $y = -\dfrac{1}{4}x + 1$ .

The tangent to y = 1/x at (2, 1/2) has slope −1/4.Open in grapher →

Instantaneous rate of change

Slope measures rate of change. The slope of a secant line is an average rate of change over an interval, and the slope of the tangent line is the instantaneous rate of change at one input.

The most familiar example is speed. If $s(t)$ is an object's position at time $t$ , then $\dfrac{s(a + h) - s(a)}{h}$ is its average velocity from time $a$ to time $a + h$ . Shrinking the time interval to nothing gives the instantaneous velocity at time $a$ :

v(a) = \lim_{h \to 0} \frac{s(a + h) - s(a)}{h}.

Worked example: Velocity of a thrown ball

A ball is thrown upward, and its height after $t$ seconds is $s(t) = 64t - 16t^2$ feet. Find its velocity at $t = 1$ .

Solution. $s(1) = 64 - 16 = 48$ . Then

s(1 + h) = 64(1 + h) - 16(1 + 2h + h^2) = 48 + 32h - 16h^2.

The difference quotient is $\dfrac{32h - 16h^2}{h} = 32 - 16h$ , which approaches $32$ . At $t = 1$ the ball is rising at $32$ feet per second.

Tip

The limit of the difference quotient is so important that calculus gives it a name: the derivative of $f$ at $a$ . Everything in this lesson is a preview of that idea.