1.11 Defining Continuity at a Point

1.11 Defining Continuity at a Point

A function is continuous at $x=a$ when its graph has no break at that input: the function is defined there, its nearby values approach a single limit, and that limit equals the actual function value.

The three-condition test at a point

The formal test is not merely “the graph looks connected.” To justify continuity at $x=a$, verify all three conditions:

A function $f$ is continuous at $x=a$ if and only if
$1.$ $f(a)$ is defined;
$2.$ $\displaystyle\lim_{x\to a}f(x)$ exists; and
$3.$ $\displaystyle\lim_{x\to a}f(x)=f(a)$.

If even one condition fails, $f$ is discontinuous at $x=a$. The strongest AP-style justification identifies the failed condition rather than simply stating that the function is “not smooth” or that its graph “has a hole.”

Worked example: a temperature sensor with a missing reading

Suppose a sensor model is

$$ T(t)=\frac{t^2-9}{t-3}, $$

where $t$ is time in minutes. At $t=3$,

$$ T(3)=\frac{3^2-9}{3-3}=\frac{0}{0}, $$

so $T(3)$ is not defined. Although algebraic simplification gives

$$ T(t)=t+3 \qquad (t\ne 3), $$

the original function still has no value at $t=3$. Therefore, $T$ is not continuous at $t=3$ because Condition 1 fails. The simplified expression describes the nearby behavior, but it does not automatically create the missing value.

Misconception check — “The limit equals the formula after cancellation, so the function is continuous.” Cancellation can reveal the limit, but continuity also requires an actual function value at the point. Always check $f(a)$ separately.

Reasoning and retrieval check

Each condition can fail on its own. In the sensor example Condition 1 failed. At a jump, where the left- and right-hand limits differ, Condition 2 fails. When the limit exists and $f(a)$ is defined but the two numbers differ—a hole with a filled point at another height—Condition 3 fails. A strong justification states $f(a)$ and $\lim_{x\to a}f(x)$ explicitly and names the condition that decides the question.

Retrieval check: Let $f(x)=x+1$ for $x\ne2$ and $f(2)=5$. Is $f$ continuous at $x=2$? Here $f(2)=5$ is defined and $\lim_{x\to2}f(x)=3$ exists, but $3\ne5$, so Condition 3 fails and $f$ is discontinuous at $x=2$.