Continuity at a Point

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Textbook Reference

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What Continuity Means

A function is continuous at a point if its graph has no break there: you could draw through that point without lifting your pen. That is the picture. The precise version turns the picture into one equation.

A function $f$ is continuous at $a$ when \[ \lim_{x \to a} f(x) = f(a). \] Read it carefully. The left side is where the function is heading as $x$ approaches $a$. The right side is the actual value at $a$. Continuity says these agree: the place the curve is aiming for is exactly the place it lands. No hole (where the curve aims somewhere the value is not), no jump (where the two sides aim at different places), no blow-up.

One equation is doing the work of three separate requirements. For $f$ to be continuous at $a$, all three must hold: the value $f(a)$ must exist, the limit must exist, and the two must be equal. Every discontinuity is one of these three failing, and naming which one fails is how you classify the break.

This pays off right away. For the functions you already know, polynomials and rational functions, continuity is automatic on their domains. That promotes the Direct Substitution Property from a convenient trick into a theorem: for a continuous function you find the limit by plugging in. Continuity is the property that makes “just substitute” a legitimate move.

Prerequisite Check

Before this lesson, make sure you can do all of the following:

• Evaluate a limit using the limit laws and direct substitution (review Properties of Limits)
• Find one-sided limits and decide whether a two-sided limit exists (review One-Sided Limits)
• Evaluate a function at a point and recognize when it is undefined (a zero denominator, for example)
• Factor a quadratic and cancel a common factor
• Read a piecewise definition and pick the right rule for a given input

If “the limit exists” versus “the function is defined” feels like the same thing, slow down. The entire topic is about cases where one holds and the other fails.

Quick Reference

Definition (continuity at a point). A function $f$ is continuous at $a$ if \[ \lim_{x \to a} f(x) = f(a). \]

The three conditions packed inside it. $f$ is continuous at $a$ exactly when all three hold:

1. $f(a)$ is defined ($a$ is in the domain).
2. $\lim_{x \to a} f(x)$ exists (both one-sided limits agree).
3. $\lim_{x \to a} f(x) = f(a)$.

If any one fails, $f$ is discontinuous at $a$.

Three kinds of discontinuity.

Which functions are continuous on their domains. Polynomials, rational functions, root functions, and trigonometric functions are all continuous at every point of their domains. Sums, differences, products, quotients (nonzero denominator), and compositions of continuous functions are continuous.

Direct Substitution Property. If $f$ is continuous at $a$, then $\lim_{x\to a} f(x) = f(a)$: find the limit by plugging in.

Key Concepts

1. The Definition and Its Three Conditions

The single equation $\lim_{x \to a} f(x) = f(a)$ demands three separate things, and pulling them apart is what makes continuity precise.

Definition. $f$ is continuous at $a$ if $\lim_{x \to a} f(x) = f(a)$. (Plain gloss: the value the curve approaches at $a$ equals the value it actually takes at $a$.)

For that equation to even make sense and then be true, all three of the following must hold:

1. $f(a)$ is defined. If $a$ is not in the domain, the right side is meaningless and continuity fails immediately.
2. $\lim_{x\to a} f(x)$ exists. The left side must be a single number, which requires the left-hand and right-hand limits to agree.
3. The two are equal. Even if both sides exist, they might be different numbers.

A continuous function has the property that a small change in $x$ produces only a small change in $f(x)$. That is the analytic meaning behind “no break in the graph.”

2. Reading Discontinuities Off a Graph

Each of the three conditions, when it fails, produces a recognizable break.

Example 1. A graph shows breaks at $x = 1$, $x = 3$, and $x = 5$. Identify which condition fails at each. (This is Stewart 1.8, Example 1.)

• At $x = 1$: the graph has a hole and $f(1)$ is not defined. Condition 1 fails.
• At $x = 3$: the value $f(3)$ is defined, but the left and right limits disagree, so the limit does not exist. Condition 2 fails.
• At $x = 5$: the value $f(5)$ is defined and the limit exists, but $\lim_{x\to 5} f(x) \neq f(5)$. Condition 3 fails.

Recap. Three different breaks, three different broken conditions. When you spot a discontinuity, the diagnostic question is always: is the value missing, is the limit missing, or do they simply disagree?

3. Classifying Discontinuities From a Formula

The same three failures have standard names. Knowing the name tells you the shape of the break and whether it can be repaired.

Example 2. Find and classify the discontinuities of each function. (This is Stewart 1.8, Example 2.)

(a) $\;f(x) = \dfrac{x^2 - x - 2}{x - 2}$

Here $f(2)$ is undefined (zero denominator). But factoring shows \[ \frac{x^2 - x - 2}{x - 2} = \frac{(x-2)(x+1)}{x-2} = x + 1 \quad (x \neq 2), \] so $\lim_{x\to 2} f(x) = 3$ exists. The limit exists but $f(2)$ does not. This is a removable discontinuity, a single hole at $(2, 3)$; defining $f(2) = 3$ would patch it.

(b) $\;f(x) = \dfrac{1}{x^2}$, with $f(0) = 1$ assigned

Now $f(0) = 1$ is defined, but $\lim_{x\to 0}\dfrac{1}{x^2} = +\infty$ does not exist (the values grow without bound). This is an infinite discontinuity, a vertical asymptote at $x = 0$.

(c) $\;f(x) = \lfloor x \rfloor$ (the greatest integer function)

At each integer $n$, the left limit is $n - 1$ and the right limit is $n$, so the one-sided limits disagree. These are jump discontinuities, one at every integer.

Recap. Removable means the limit exists and only the value is wrong (or missing); the graph has a single hole you could fill. Jump means the two sides head to different heights. Infinite means the function blows up. The classification is exactly which of the three conditions broke.

Important: a removable discontinuity still makes the function discontinuous. “Removable” describes a hole that could be patched by redefining one point, not a function that is already continuous. Until you actually redefine $f(2) = 3$ in part (a), the function is discontinuous at $2$. The word names the repair that is possible, not a repair already done.

4. Continuity From One Side

At an endpoint of a domain, or at a jump, it helps to talk about continuity from a single direction.

Definition. $f$ is continuous from the right at $a$ if $\lim_{x\to a^+} f(x) = f(a)$, and continuous from the left at $a$ if $\lim_{x\to a^-} f(x) = f(a)$.

A function is continuous at $a$ in the full sense exactly when it is continuous from both sides.

Example 3. Show that $f(x) = \lfloor x \rfloor$ is continuous from the right but not from the left at each integer $n$. (This is Stewart 1.8, Example 3.)

From the right, $\lfloor x \rfloor = n$ for $n \leq x < n+1$, so $\lim_{x\to n^+}\lfloor x\rfloor = n = f(n)$: continuous from the right. From the left, $\lfloor x \rfloor = n - 1$ for $n-1 \leq x < n$, so $\lim_{x\to n^-}\lfloor x\rfloor = n - 1 \neq f(n)$: not continuous from the left.

Recap. One-sided continuity is the right tool at a step or a domain edge. The greatest-integer function steps up exactly at each integer, so it catches the value from the right but not from the left.

5. The Functions That Are Always Continuous

Verifying continuity from the definition every time would be tedious. The good news is that the standard functions are continuous wherever they are defined, and continuity is preserved by every basic combination.

Theorem. If $f$ and $g$ are continuous at $a$, then so are $f + g$, $f - g$, $cf$, $fg$, and $f/g$ (provided $g(a) \neq 0$). Each follows from the matching limit law.

Theorem. Polynomials are continuous on all of $\mathbb{R}$. Rational functions, root functions, and trigonometric functions are continuous at every point of their domains.

Theorem (composition). If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then $f \circ g$ is continuous at $a$: a continuous function of a continuous function is continuous.

Together these let you certify continuity by inspection. They also justify the Direct Substitution Property: for a function built from continuous pieces, $\lim_{x\to a} f(x) = f(a)$.

Example 4. Evaluate $\displaystyle\lim_{x\to \pi}\frac{\sin x}{2 + \cos x}$. (This is Stewart 1.8, Example 7.)

Goal. Show the function is continuous at $\pi$, then substitute.

The numerator $\sin x$ is continuous. The denominator $2 + \cos x$ is a sum of continuous functions and is never zero (since $\cos x \geq -1$, so $2 + \cos x \geq 1 > 0$). By the quotient theorem the whole function is continuous everywhere, so substitution is valid: \[ \lim_{x\to \pi}\frac{\sin x}{2 + \cos x} = \frac{\sin \pi}{2 + \cos \pi} = \frac{0}{2 - 1} = 0. \]

Boxed answer: $0$.

Recap. Once you recognize a function as a combination of continuous pieces with a nonzero denominator, the limit is just the value. That recognition is faster than any algebraic technique and is the everyday use of continuity.

6. The Intermediate Value Theorem

Continuity on a whole interval has a consequence: a continuous function cannot skip values.

The Intermediate Value Theorem. If $f$ is continuous on $[a, b]$ and $N$ is any number between $f(a)$ and $f(b)$, then there is a number $c$ in $(a, b)$ with $f(c) = N$. (Plain gloss: a graph drawn without lifting the pen cannot jump over a horizontal line between its endpoints; it must cross.)

The standard use is to prove an equation has a solution.

Example 5. Show that $4x^3 - 6x^2 + 3x - 2 = 0$ has a solution between $1$ and $2$. (This is Stewart 1.8, Example 9.)

Goal. Find a sign change, then invoke the theorem with $N = 0$.

Let $f(x) = 4x^3 - 6x^2 + 3x - 2$, a polynomial, hence continuous. Evaluate the endpoints: \[ f(1) = 4 - 6 + 3 - 2 = -1 < 0, \qquad f(2) = 32 - 24 + 6 - 2 = 12 > 0. \] Since $0$ lies between $f(1)$ and $f(2)$, the Intermediate Value Theorem guarantees a $c$ in $(1, 2)$ with $f(c) = 0$.

Boxed answer: a solution exists in $(1, 2)$. Narrowing further, $f(1.2) = -0.128 < 0$ and $f(1.3) = 0.548 > 0$, so a root lies in $(1.2, 1.3)$.

Recap. A sign change in a continuous function forces a root between. This is exactly how a computer finds roots by bisection: check the midpoint’s sign and keep the half that still straddles zero. Continuity is the guarantee that the method works.

Common Errors Summary

Leveled Practice

Level 1 -- Direct Application

Problem 1. Use the three-condition test to decide whether $f(x) = \dfrac{x^2 + 2x + 17}{x^2 - 1}$ is continuous at $x = 0$.

Problem 2. Classify the discontinuity of $f(x) = \dfrac{x^2 - 7x + 12}{x - 3}$ at $x = 3$, and state the value that would remove it.

Problem 3. Classify the discontinuity of $f(x) = \dfrac{1}{x + 2}$ at $x = -2$.

Level 2 -- Multiple Steps

Problem 4. State the intervals on which $g(x) = \dfrac{x^2 + 2x + 17}{x^2 - 1}$ is continuous.

Problem 5. Determine whether $f(x) = \begin{cases} 1 - x^2 & \text{if } x \leq 1 \\ \sqrt{x - 1} & \text{if } x > 1 \end{cases}$ is continuous at $x = 1$.

Problem 6. For what value of the constant $c$ is $f(x) = \begin{cases} cx^2 + 2 & \text{if } x < 2 \\ x^3 - cx & \text{if } x \geq 2 \end{cases}$ continuous at $x = 2$?

Level 3 -- Deeper Problems

Problem 7. Use the Intermediate Value Theorem to show that the equation $\cos x = x$ has a solution in $(0, 1)$.

Problem 8. Where is $h(x) = \sin(x^2)$ continuous, and why?

Problem 9. A Tibetan monk leaves the monastery at 7:00 AM and reaches the mountaintop at 7:00 PM. The next morning he leaves the top at 7:00 AM along the same path and reaches the monastery at 7:00 PM. Use the Intermediate Value Theorem to show there is a point on the path he passes at exactly the same time of day on both trips.

Common Misconceptions

Mastery Checklist

You have mastered this skill when you can do all of the following without referring to notes:

• State the definition of continuity at a point as $\lim_{x\to a} f(x) = f(a)$
• List the three conditions hidden in that equation and test a function against each
• Classify a discontinuity as removable, jump, or infinite, and say which condition failed
• Explain why a removable discontinuity is still a discontinuity until the point is redefined
• Use continuity from the left and right to handle endpoints and steps
• Name the function families that are continuous on their domains and use them to certify continuity by inspection
• Apply the Direct Substitution Property to evaluate a limit of a continuous function
• Use the Intermediate Value Theorem to prove an equation has a solution in an interval

Mental Model

Think of continuity as the curve landing where it was aiming.

As $x$ approaches $a$, the function is heading somewhere; that destination is the limit. The function also has an actual value at $a$. Continuity is the statement that these coincide: the curve aims at $f(a)$ and lands on $f(a)$, with no hole, no jump, no blow-up.

Three things follow from the picture:

• A break is one of three failures. Either there is no value to land on (a hole, condition 1), or there is no single destination because the two sides aim differently (a jump, condition 2), or the value and the destination disagree (condition 3). Naming the failure names the discontinuity.
• “No break” means you can substitute. For a function built from continuous pieces, the destination is the value, so evaluating a limit is just plugging in. This is why the Direct Substitution Property works and why continuity is the property that makes calculus computable.
• Continuity forbids skipping. A curve you can draw without lifting your pen cannot hop over a horizontal line strung between its endpoints; it must cross. That is the Intermediate Value Theorem, and it is why a sign change guarantees a root.

Connections

Within Functions and Limits (Chapter 1)

• Properties of limits: Continuity is built directly on the limit laws; each continuity-of-combinations theorem is the matching limit law restated. The Direct Substitution Property is continuity in one sentence.
• One-sided limits: Jump discontinuities and continuity at an endpoint are diagnosed with one-sided limits. A point is continuous exactly when it is continuous from both sides.
• Continuity on an interval and the Intermediate Value Theorem: Continuity at every point of an interval is continuity on the interval, and that is the hypothesis the Intermediate Value Theorem needs. Both have their own nodes that build on this one.

Toward later calculus (MATH161 and beyond)

• Differentiability: Differentiable implies continuous; a function with a corner or a break cannot have a derivative there. Continuity is the weaker, prior condition every differentiable function satisfies.
• The Extreme Value Theorem and the Mean Value Theorem: Both require continuity on a closed interval. The guarantee that a continuous function on $[a, b]$ attains a maximum and minimum, used inside Rolle’s Theorem, rests on the continuity developed here.
• Integrability: Every continuous function on a closed interval is integrable, so the definite integral of a continuous function always exists. Continuity is the standard sufficient condition that makes the limit of Riemann sums converge.

Audience Notes

For students who find math intimidating: You can answer almost every continuity question with one checklist. At the point in question, ask: (1) does the function have a value there, (2) does the limit exist there, (3) do they match? If all three are yes, it is continuous; if any is no, that is your discontinuity, and which one failed names the type. For familiar functions like polynomials, the answer is simply yes everywhere.

For career-focused students: The Intermediate Value Theorem is the mathematical license for root-finding by bisection, a method used in every numerical library: bracket a sign change, repeatedly halve the interval, and the root is trapped. Continuity is also what lets a graphics program “connect the dots” between computed points, assuming the curve takes every value in between.

For gifted and curious students: Investigate the function that equals $0$ at every rational number and $1$ at every irrational number. It is discontinuous everywhere, yet a close cousin (the Thomae function) is continuous at every irrational and discontinuous at every rational. These examples show how much subtler continuity becomes once you leave the familiar formula-defined functions.

For PhD-track students: The Intermediate Value Theorem is equivalent to the completeness of the real numbers; it fails over the rationals (where $x^2 = 2$ has no solution despite a sign change). The “small change in input gives small change in output” phrasing is pointwise continuity, which is weaker than uniform continuity, and the distinction, invisible on a closed bounded interval by the Heine-Cantor theorem, becomes essential on open or unbounded domains.

Back to Functions and Limits | Related: Properties of Limits | Next: The Intermediate Value Theorem