Math Core

Calculus is the mathematics of change. Its two central questions, "how fast is something changing at a single instant?" and "how much total change accumulates?", both lead to the same tool: the limit. Before you can define a derivative or an integral, you need a precise way to talk about what a function is approaching, even at a point where you cannot simply plug in.

Why calculus needs limits

Suppose a ball's height is $s(t)$ feet after $t$ seconds. The average velocity from $t = 1$ to $t = 1 + h$ is

\frac{s(1 + h) - s(1)}{h}.

To get the velocity at the instant $t = 1$ , you would like to set $h = 0$ , but then the fraction becomes $\dfrac{0}{0}$ , which is meaningless. What you can do is make $h$ smaller and smaller ( $0.1$ , $0.01$ , $0.001$ , …) and watch the average velocities. If they settle toward one number, that number is the instantaneous velocity. That "settling toward" is exactly what a limit captures.

The idea of a limit

Consider

f(x) = \frac{x^2 - 4}{x - 2}.

The function is undefined at $x = 2$ (the denominator is zero). But for every other $x$ you can factor and cancel:

f(x) = \frac{(x - 2)(x + 2)}{x - 2} = x + 2, \qquad x \ne 2.

So the graph of $f$ is the line $y = x + 2$ with a single point missing, a hole at $(2, 4)$ .

The graph of f(x) = (x² − 4)/(x − 2): the line y = x + 2 with a hole at (2, 4).Open in grapher →

As $x$ gets close to $2$ from either side, $f(x)$ gets close to $4$ . We write

\lim_{x \to 2} f(x) = 4,

read "the limit of $f(x)$ as $x$ approaches $2$ equals $4$ ." Notice that $f(2)$ does not exist, yet the limit does. The limit describes the behavior near $2$ , never the value at $2$ .

Definition

Limit (informal)

We write $\displaystyle \lim_{x \to c} f(x) = L$ if the values of $f(x)$ can be made as close to $L$ as we like by taking $x$ sufficiently close to $c$ (on either side of $c$ ), but not equal to $c$ .

Three things to take from this definition:

$x$ never equals $c$ . What happens exactly at $x = c$ (whether $f(c)$ is defined, and what it equals) has no effect on the limit.
Both sides matter. $x$ can approach $c$ from the left (values less than $c$ ) and from the right (values greater than $c$ ).
$L$ must be a single real number. If the outputs head toward two different numbers, or grow without bound, or never settle down, the limit does not exist.

One-sided limits

Sometimes a function behaves differently on the two sides of $c$ . For this reason we also define one-sided limits:

$\displaystyle \lim_{x \to c^-} f(x) = L$ means $f(x)$ approaches $L$ as $x$ approaches $c$ from the left ( $x < c$ ).
$\displaystyle \lim_{x \to c^+} f(x) = L$ means $f(x)$ approaches $L$ as $x$ approaches $c$ from the right ( $x > c$ ).

The small minus or plus sign is a superscript on $c$ . It describes the direction of approach, not the sign of the number.

When a two-sided limit exists

\lim_{x \to c} f(x) = L \quad \text{if and only if} \quad \lim_{x \to c^-} f(x) = L \ \text{ and } \ \lim_{x \to c^+} f(x) = L.

If the one-sided limits are different, or if either one fails to exist, then $\displaystyle \lim_{x \to c} f(x)$ does not exist (DNE).

Three ways a limit can fail to exist

On the AP exam you will meet three standard situations in which $\displaystyle \lim_{x \to c} f(x)$ does not exist.

Behavior near $c$	Example	Why the limit fails
Jump: the two sides approach different values	$\dfrac{\lvert x \rvert}{x}$ at $c = 0$	left limit is $-1$ , right limit is $1$
Unbounded: the outputs grow without bound	$\dfrac{1}{x^2}$ at $c = 0$	outputs increase past every number
Oscillation: the outputs never settle	$\sin\left(\dfrac{1}{x}\right)$ at $c = 0$	outputs swing between $-1$ and $1$ forever

For unbounded behavior we often write $\displaystyle \lim_{x \to 0} \frac{1}{x^2} = \infty$ . This notation is a precise description of how the limit fails: the outputs increase without bound. It does not mean the limit equals a number called infinity, and the limit still does not exist as a real number.

Common mistake

Do not confuse $\displaystyle \lim_{x \to c} f(x)$ with $f(c)$ . A function can have a limit at $c$ with $f(c)$ undefined, or with $f(c)$ equal to some completely different number. Always ask "what are the outputs approaching?", not "what is the output at $c$ ?"

Worked example: A limit at a hole

Let $g(x) = \dfrac{x^2 + x - 12}{x - 3}$ . Find $\displaystyle \lim_{x \to 3} g(x)$ and $g(3)$ .

Solution. At $x = 3$ the denominator is $0$ , so $g(3)$ is undefined. For $x \ne 3$ ,

g(x) = \frac{(x + 4)(x - 3)}{x - 3} = x + 4.

As $x \to 3$ , $x + 4 \to 7$ . So $\displaystyle \lim_{x \to 3} g(x) = 7$ , even though $g(3)$ does not exist.

Worked example: One-sided limits of a piecewise function

Let

p(x) = \begin{cases} 2x - 1, & x < 1 \\ 4, & x = 1 \\ x^2 + 2, & x > 1. \end{cases}

Find $\displaystyle \lim_{x \to 1^-} p(x)$ , $\displaystyle \lim_{x \to 1^+} p(x)$ , $\displaystyle \lim_{x \to 1} p(x)$ , and $p(1)$ .

Solution. For $x$ slightly less than $1$ , $p(x) = 2x - 1$ , which approaches $2(1) - 1 = 1$ . For $x$ slightly greater than $1$ , $p(x) = x^2 + 2$ , which approaches $1 + 2 = 3$ .

\lim_{x \to 1^-} p(x) = 1, \qquad \lim_{x \to 1^+} p(x) = 3.

The one-sided limits differ, so $\displaystyle \lim_{x \to 1} p(x)$ does not exist. Separately, $p(1) = 4$ from the middle line of the definition.

Worked example: Reading a limit from a graph

The graph of $h$ is shown. Find $\displaystyle \lim_{x \to 2} h(x)$ and $h(2)$ .

The graph of h. The open circle is at (2, 3); the filled dot is at (2, 1).Open in grapher →

Solution. Tracing the graph toward $x = 2$ from the left, the heights approach $3$ . Tracing from the right, the heights also approach $3$ . Both one-sided limits equal $3$ , so $\displaystyle \lim_{x \to 2} h(x) = 3$ . The filled dot shows that the actual value is $h(2) = 1$ . Again, the limit and the function value are different numbers.

Tip

When you read a limit from a graph, cover the vertical line $x = c$ with your finger. The limit depends only on what you can still see.

Practice

Practice 1

Let $f(x) = \dfrac{x^2 - 9}{x - 3}$ . Find $\displaystyle \lim_{x \to 3} f(x)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

The graph of $h$ from the third example is shown again. Which statement is true?

The graph of h. The open circle is at (2, 3); the filled dot is at (2, 1).Open in grapher →

Practice 3

Let $k(x) = \begin{cases} x^2, & x < 1 \\ 2x + 3, & x \ge 1. \end{cases}$ Find $\displaystyle \lim_{x \to 1^-} k(x)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

For the function $k$ in the previous problem, what is $\displaystyle \lim_{x \to 1} k(x)$ ?

Practice 5

If $\displaystyle \lim_{x \to 4} f(x) = 7$ , which of the following must be true?

Practice 6

Which best describes $\displaystyle \lim_{x \to 0} \frac{1}{x^2}$ ?

Practice 7

Let $r(x) = \dfrac{x^2 + 7x + 10}{x + 2}$ for $x \ne -2$ and $r(-2) = 8$ . Find $\displaystyle \lim_{x \to -2} r(x)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.