Calculus I · Limits and Continuity · lesson

Continuity at a Point

Visual study stop

Read the picture before the symbols

Pause at each graph long enough to say what the input is doing, what the output is doing, and which feature supports the next mathematical decision.

Parabola with an open point at (2, 5) and a filled point at (2, 9). — **Test the three continuity conditions as separate claims**
The nearby parabola supplies the limit while the isolated filled point supplies the function value. Continuity requires both to exist and to land at the same height.

Concept

Learning objectives

Use the three-part continuity test at a point; distinguish continuity from merely having a limit; explain the graphical meaning of continuity.

Continuity and the Intermediate Value Theorem

Continuity at a Point

A limit describes nearby behavior. Continuity connects that nearby behavior to the function's actual value.

Definition

Continuity at $x=a$

A function $f$ is continuous at $x=a$ if all three conditions hold:

• $f(a)$ is defined; • $\displaystyle\lim_{x\to a}f(x)$ exists; • $\displaystyle\lim_{x\to a}f(x)=f(a)$ .

Equivalently,

\boxed{\lim_{x\to a}f(x)=f(a)}.

Concept

Continuity means the graph, the nearby trend, and the actual dot all agree at the point. You approach one height, and the function is actually located at that height.

Guided walkthrough

A continuous line

Is $f(x)=2x+1$ continuous at $x=3$ ?

Show worked solution

Check the three conditions.

1. The function value exists:

f(3)=2(3)+1=7.

2. The limit exists:

\lim_{x\to3}(2x+1)=7.

3. They agree:

\lim_{x\to3}f(x)=f(3)=7.

Therefore, $f$ is continuous at $3$ .

Worked example

A limit exists but continuity fails

Let

g(x)= \begin{cases} x^2+1,&x\ne2,\\ 9,&x=2. \end{cases}

Is $g$ continuous at $2$ ?

Show worked solution

Condition 1: $g(2)=9$ , so the function value exists.

Condition 2: Nearby values use $x^2+1$ , so

\lim_{x\to2}g(x)=5.

The limit exists.

Condition 3: Compare:

\lim_{x\to2}g(x)=5\ne9=g(2).

Therefore, $g$ is not continuous at $2$ .

This discontinuity is removable. Redefining $g(2)=5$ would repair it.

Worked example

Continuity fails because no two-sided limit exists

Let

h(x)= \begin{cases} x+1,&x<2,\\ 6-x,&x\ge2. \end{cases}

Is $h$ continuous at $2$ ?

Show worked solution

The function value exists:

h(2)=6-2=4.

But

\lim_{x\to2^-}h(x)=3,

while

\lim_{x\to2^+}h(x)=4.

The one-sided limits disagree, so the two-sided limit does not exist. Condition 2 fails. Therefore, $h$ is not continuous at $2$ .

Exam note

Do not use only the slogan "draw without lifting your pencil." It is a helpful picture, not a complete test. At endpoints, one-sided continuity is allowed; on disconnected domains, a graph can be continuous at every point of its domain even though you cannot draw all pieces in one stroke.

After the explanation

Use the section idea

Reading lens

Do the limit, the function value, and the surrounding domain fit together at the point or across the interval?

Mental model

Continuity is a three-part agreement: the value exists, the two-sided limit exists, and those two quantities are equal.

Decision

At a point, test the three conditions in order; on an interval, check the domain and endpoints before invoking any continuity theorem.

Common trap

A sign change supports the Intermediate Value Theorem only when continuity holds on the entire closed interval, and it does not prove uniqueness.

Check yourself

You are ready to continue when you can classify a break, decide whether one value can repair it, and state every IVT hypothesis aloud.

Continue

Use one focused companion

Quick explanationWhat continuity at a point really requires ReferenceContinuity: Meaning, Notation, and Example

Source & rights

Original instruction with traceable references.

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