Free AP Calculus AB Limits and Continuity

AP Calculus AB · Limits and Continuity · free lesson

Introducing Calculus; Defining Limits and Limit Notation

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

What calculus is about

Calculus studies change. Its first question is how fast something is changing at one instant: the speed of a car at the moment it passes a sign, the rate at which a tank is filling at noon. Its second question, later in the course, is how much has built up when the rate keeps changing. Both questions are answered with one idea, the limit.

An average rate over a shorter and shorter time

A stone is dropped from a high bridge. After $t$ seconds it has fallen $s(t) = 16t^2$ feet. Over a time interval, its average speed is the distance fallen divided by the time taken. From $t = 2$ to $t = 2 + h$ the average speed, in feet per second, is

\begin{aligned} \frac{s(2 + h) - s(2)}{h} &= \frac{16(2 + h)^2 - 16 \cdot 2^2}{h} \end{aligned}

The engine works it out for intervals that get shorter and shorter, after the moment $t = 2$ (positive $h$ ) and before it (negative $h$ ):

Interval	$h$	Average speed (ft/s)
$[2, 3]$	$1$	$80$
$[2, 2.1]$	$0.1$	$65.6$
$[2, 2.01]$	$0.01$	$64.16$
$[2, 2.001]$	$0.001$	$64.016$
$[1.999, 2]$	$-0.001$	$63.984$
$[1.99, 2]$	$-0.01$	$63.84$
$[1.9, 2]$	$-0.1$	$62.4$

From both sides the averages close in on one number, $64$ . That number is the stone's speed at the instant $t = 2$ : $64$ feet per second. In the picture each average speed is the slope of a secant line through two points of the graph. As $h$ shrinks, the second point slides toward the first and the secant line turns toward the line that just touches the curve at $t = 2$ .

Warning.

You cannot simply put $h = 0$ . An interval of length 0 gives $\frac{s(2) - s(2)}{0} = \frac{0}{0}$ , which has no value: no distance is fallen and no time passes. The speed at an instant is not an average; it is the number the averages approach.

The limit and its notation

Definition.

If the values $f(x)$ can be made as close to a number $L$ as we like by taking $x$ close enough to $c$ (on both sides, but not equal to $c$ ), then $L$ is the limit of $f(x)$ as $x$ approaches $c$ . It is written

\begin{aligned} \lim_{x \to c} f(x) &= L \end{aligned}

Read it aloud as "the limit of $f$ of $x$ , as $x$ approaches $c$ , is $L$ ". The arrow and the word "lim" are part of the statement: on the AP exam, writing only $f(x) = L$ says something else. (The exact form of "as close as we like", with $\varepsilon$ and $\delta$ , is not asked on the AP exam.)

The function $f(x) = \frac{\sin x}{x}$ has no value at $x = 0$ , since there it is $\frac{\sin 0}{0} = \frac{0}{0}$ . Its graph has a hole there. Drag the slider to bring the two red points toward 0 from both sides, and watch their height.

At distance $1$ from 0 the height is $0.841471$ ; at $0.1$ it is $0.998334$ ; at $0.01$ it is $0.999983$ . The heights close in on $1$ , so $\lim_{x \to 0} \frac{\sin x}{x} = {1}$ , although $f(0)$ does not exist.

The limit is not the value

A limit is about the values near $c$ , never the value at $c$ . Below, $g(x) = x + 1$ except at $x = 2$ , where $g(2) = 1$ (the filled dot). Near 2 the graph runs toward the hole.

So $\lim_{x \to 2} g(x) = {3}$ while $g(2) = 1$ . A limit may equal the value, may differ from it, or may exist where there is no value at all.

One-sided limits

Sometimes the two sides disagree. $\lim_{x \to c^-} f(x)$ is the limit from the left (values of $x$ less than $c$ ); $\lim_{x \to c^+} f(x)$ is the limit from the right (values greater than $c$ ). Here $p(x) = 3 - x$ for $x < 1$ and $p(x) = x + 2$ for $x \ge 1$ :

From the left the graph runs to height $2$ , so $\lim_{x \to 1^-} p(x) = {2}$ . From the right it runs to $3$ , so $\lim_{x \to 1^+} p(x) = {3}$ . The two sides differ, so $\lim_{x \to 1} p(x)$ does not exist.

Rule.

$\lim_{x \to c} f(x) = L$ exactly when both one-sided limits exist and both equal $L$ .

When the values grow without bound

Near $x = 2$ the values of $\frac{1}{(x - 2)^2}$ grow past every number, from both sides. There is no number they approach, so the limit does not exist; but the way it fails is written

\begin{aligned} \lim_{x \to 2} \frac{1}{(x - 2)^2} &= \infty \end{aligned}

Note.

On the AP exam, use $= \infty$ (or $= -\infty$ ) when the values grow without bound in one direction, and say "does not exist" when the two sides disagree or the values never settle. The usual slips: taking $f(c)$ for the limit, mixing up $c^-$ and $c^+$ , and leaving out "lim" or the arrow.

Worked examples

Each example is one kind of question the AP exam asks about rates over shrinking intervals and about the notation of limits. Every number in the working is found by the engine.

Example.

1. Average rates that close in on a rate at an instant. A tank is filling with water. After $t$ minutes it holds $V(t) = t^{3}$ gallons. Find the average rate at which the water comes in over $[2, 2 + h]$ for $h = 0.1$ , $0.01$ , $0.001$ and $h = -0.001$ , $-0.01$ , $-0.1$ . What value do the rates approach?

The average rate over $[2, 2 + h]$ is the change in volume divided by the change in time, $\frac{V(2 + h) - V(2)}{h}$ gallons per minute. A negative $h$ is an interval just before $t = 2$ .

$h$	$0.1$	$0.01$	$0.001$	$-0.001$	$-0.01$	$-0.1$
Average rate (gal/min)	$12.61$	$12.0601$	$12.006001$	$11.994001$	$11.9401$	$11.41$

From both sides the rates close in on $12$ . So at the instant $t = 2$ the water is coming in at $12$ gallons per minute. In symbols,

\begin{aligned} \lim_{h \to 0} \frac{V(2 + h) - V(2)}{h} &= 12 \end{aligned}

The picture shows the secant lines for $h = 1$ and $h = \frac{1}{2}$ ; their slopes are the average rates, and they get closer to the slope of the curve at $t = 2$ .

Check: the secant slopes are $19$ for $h = 1$ and $15.25$ for $h = \frac{1}{2}$ . With the rates in the table they fall toward $12$ as $h$ shrinks.

Example.

2. Why an interval of length 0 does not work. A student wants the rate at the instant $t = 2$ for the tank and writes $\frac{V(2) - V(2)}{2 - 2}$ . What is wrong?

The top is $V(2) - V(2) = 0$ and the bottom is $2 - 2 = 0$ . A quotient of zero by zero has no value, so the average rate is undefined for an interval of length 0. The rate at an instant is not found by putting the two times together; it is the limit of the average rates as the interval shrinks, as in Example 1.

Example.

3. From words to notation and back. Write each sentence as a limit statement.

As $x$ approaches 4 from the left, $f(x)$ approaches 7.
As $x$ approaches $-1$ from the right, $g(x)$ approaches 0.
As $t$ approaches 3, from either side, $h(t)$ approaches $-2$ .
As $x$ approaches 5, $k(x)$ grows without bound.

"From the left" means through values less than the number, written with a minus sign up high; "from the right" means through greater values, with a plus sign. No sign means both sides. Growing without bound is written $= \infty$ .

$\lim_{x \to 4^-} f(x) = 7$
$\lim_{x \to -1^+} g(x) = 0$
$\lim_{t \to 3} h(t) = -2$
$\lim_{x \to 5} k(x) = \infty$

Read backwards, $\lim_{x \to 4^-} f(x) = 7$ says nothing about $x$ greater than 4 and nothing about $f(4)$ itself.

Example.

4. What a limit statement does and does not tell you. It is known that $\lim_{x \to 3} f(x) = 5$ . Must $f(3)$ be 5? Must $f(3)$ exist?

No to both. The statement is about values of $f$ near 3. Here are three functions, each with limit 5 at 3: the first has $f(3) = 5$ , the second has a hole at 3 and no value there, the third has a hole and the value 2 at 3.

In all three pictures the graph runs toward height $5$ as $x$ approaches 3, so in all three $\lim_{x \to 3} f(x) = {5}$ . Only the first has $f(3) = 5$ . A limit statement alone never tells you $f(c)$ .

Example.

5. One side up, one side down. Describe the behavior of $q(x) = \frac{1}{x - 1}$ near $x = 1$ with limit notation.

Just left of 1, $x - 1$ is a small negative number, so $q(x)$ is a large negative number; just right of 1 it is a large positive number. The engine finds the two one-sided limits:

$x$	$0.9$	$0.99$	$1.01$	$1.1$
$q(x)$	$-10$	$-100$	$100$	$10$

On the left the values fall without bound and on the right they rise without bound; the engine's one-sided limits say the same. So

\begin{aligned} \lim_{x \to 1^-} \frac{1}{x - 1} &= -\infty \\ \lim_{x \to 1^+} \frac{1}{x - 1} &= \infty \end{aligned}

The two-sided limit $\lim_{x \to 1} \frac{1}{x - 1}$ does not exist, and it cannot be written $= \infty$ either: the two sides go in opposite directions. Compare $\frac{1}{(x - 1)^2}$ , which grows without bound on both sides, so its two-sided limit is written $= \infty$ .

Practice

Answer AP-style questions on introducing calculus; defining limits and limit notation. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise introducing calculus; defining limits and limit notation with new numbers