Continuity at a Point — AP® Calculus AB Notes

The last topic described breaks in a graph. This topic turns "no break" into a precise test you can apply to a graph, a table or a formula, and write up as a justification. The test has three conditions, and every one of them matters.

The definition

A function $f$ is continuous at $x = c$ when all three of these hold:

$f(c)$ is defined. (There is a point on the graph at $x = c$ .)
$\lim_{x \to c} f(x)$ exists. (The curve arrives at one height from both sides.)
$\lim_{x \to c} f(x)$ $= f(c)$ . (The point is exactly where the curve arrives.)

Four graphs near x = c. First: an open circle and no closed dot, so f(c) is not defined and condition 1 fails. Second: a jump, so the limit at c does not exist and condition 2 fails. Third: a hole with the closed dot f(c) placed above it, so the limit exists but does not equal f(c) and condition 3 fails. Fourth: an unbroken curve through a closed dot, where all three conditions hold. — **Figure 1.** Each condition can fail while the earlier ones hold. LevlPrep original diagram.

Each condition catches a different kind of break. Check them in order: the first one that fails is your reason.

First condition to fail	What the graph shows	Kind of discontinuity
1: $f(c)$ undefined	no point at $x = c$	removable (a hole) or infinite, depending on the limit
2: the limit does not exist	the two sides arrive at different heights, or grow without bound	jump or infinite
3: limit $\ne f(c)$	a hole with the point placed elsewhere	removable

Checking a formula

Worked example 1. Let $f(x) = \frac{x^2 - 4}{x - 2}$ for $x \ne 2$ and $f(2) = 5$ . Is $f$ continuous at $x = 2$ ?

Condition 1: $f(2) = 5$ , so it is defined.

Condition 2: for $x \ne 2$ , $f(x)$ $= \frac{(x - 2)(x + 2)}{x - 2}$ $= x$ $+ 2$ , so $\lim_{x \to 2} f(x) = 4$ . The limit exists.

Condition 3: $4 \ne 5$ . It fails.

Conclusion: $f$ is not continuous at 2, because $\lim_{x \to 2} f(x)$ $= 4 \ne 5$ $= f(2)$ . Had the rule been $f(2) = 4$ , all three conditions would hold.

Checking a piecewise function

At a break point, the one-sided limits come from different pieces. Find each one, then compare.

Worked example 2. Let $g(x) = x^2 - 1$ for $x < 2$ , $g(2) = 3$ , and $g(x) = 4x - 5$ for $x > 2$ . Show that $g$ is continuous at $x = 2$ .

Condition 1: $g(2) = 3$ .

Condition 2: $\lim_{x \to 2^-} g(x)$ $= 2^2 - 1$ $= 3$ and $\lim_{x \to 2^+} g(x)$ $= 4(2) - 5$ $= 3$ . The one-sided limits are equal, so $\lim_{x \to 2} g(x) = 3$ .

Condition 3: $\lim_{x \to 2} g(x)$ $= 3$ $= g(2)$ .

All three hold, so $g$ is continuous at $x = 2$ .

Notice the write-up. Each condition gets its own sentence with its own number. "The pieces match" or "there is no break" is a conclusion, not a justification.

Reading a graph or a table

On a graph, ask three questions at $x = c$ . Is there a closed dot? Does the curve arrive at the same height from both sides? Is the dot at that height? A closed dot at the end of one piece (a jump) passes the first question and fails the second. A dot floating above a hole passes the first two and fails the third.

A table can suggest continuity: if $f(c) = 2$ is listed and the values on both sides close in on 2, continuity at $c$ is likely. As always, a table estimates the limit; it does not prove it.

Why continuity matters

If $f$ is continuous at $c$ , then $\lim_{x \to c} f(x)$ $= f(c)$ . That single fact is behind direct substitution: polynomials are continuous everywhere, so their limits are their values. It also works in reverse. If you know a function is continuous at 4 and you can find its limit there, you know its value.

Worked example 3. $f$ is continuous at $x = 4$ and $\lim_{x \to 4} (f(x) + 2x)$ $= 13$ . Find $f(4)$ .

By the limit laws, $\lim_{x \to 4} f(x)$ $= 13 - \lim_{x \to 4} 2x$ $= 13 - 8$ $= 5$ .

Continuity at 4 gives $f(4)$ $= \lim_{x \to 4} f(x)$ $= 5$ .

Later in the course, big theorems start with "if $f$ is continuous". Each time, you will need to say why the function is continuous before you use the theorem. This definition is where that habit begins.