Continuity

Understand what it means for a function to have no breaks or jumps

You have probably heard the advice to “draw a graph without lifting your pencil.” That image captures the essence of continuity perfectly. A continuous function flows smoothly from point to point, with no sudden jumps, no holes, and no places where the curve shoots off to infinity. But you already have strong intuitions about continuity from everyday life. When you watch a car drive down a highway, you expect its position to change smoothly over time. It does not teleport from one spot to another. When the temperature rises from morning to afternoon, it passes through every degree in between. It does not skip from 60 degrees directly to 75 degrees. These are continuous processes, and mathematically, they are described by continuous functions.

What makes continuity such a fundamental concept in calculus is that continuous functions behave predictably. They have no surprises. If you know a continuous function equals 2 somewhere and equals 5 somewhere else, you can be certain it hit 3 and 4 along the way. This simple observation, called the Intermediate Value Theorem, has profound consequences. It lets us prove equations have solutions without actually solving them. It guarantees that physical processes behave smoothly. Continuity is the bridge between the intuitive world of smooth curves and the precise world of limits.

Core Concepts

The Intuitive Idea

Think of continuity as “connectedness.” A continuous function has a graph that is a single, unbroken curve. You could trace it with your finger without ever having to jump from one point to another.

Consider the function $f(x) = x^2$. If you plot it, you get a smooth parabola. Starting from any point on the curve, you can move to any other point by sliding along the curve itself. There are no gaps, no breaks, no places where you would have to lift your pencil and put it down somewhere else.

Now contrast this with a function like the one that equals 1 for all positive numbers and 0 for all non-positive numbers. At $x = 0$, something dramatic happens: the function suddenly jumps from 0 to 1. That jump is a discontinuity.

The Formal Definition

The intuitive “pencil” test is helpful, but mathematics demands precision. Here is the formal definition of continuity at a point:

A function $f$ is continuous at $x = a$ if and only if:

$$\lim_{x \to a} f(x) = f(a)$$

This elegant equation packs three requirements into one. For the equation to even make sense, three things must be true:

  1. $f(a)$ must exist. The function must be defined at the point $a$. You cannot be continuous at a place where you do not exist.

  2. $\lim_{x \to a} f(x)$ must exist. The limit of $f(x)$ as $x$ approaches $a$ must exist. This means the function must be “heading somewhere” as you approach $a$ from both sides, and it must be heading to the same place from both directions.

  3. The limit must equal the function value. The place the function is heading as you approach $a$ must actually be where the function ends up at $a$. No surprises.

If any one of these conditions fails, the function is discontinuous at $a$. Understanding which condition fails tells you what type of discontinuity you have.

The Three Conditions, Unpacked

Let us examine each condition more carefully.

Condition 1: $f(a)$ is defined. This is the most basic requirement. If we are asking whether $f$ is continuous at $x = 3$, the function $f$ must actually have a value at $x = 3$. A function like $g(x) = \frac{1}{x-3}$ is not defined at $x = 3$, so it cannot possibly be continuous there.

Condition 2: $\lim_{x \to a} f(x)$ exists. For this limit to exist, the left-hand limit and right-hand limit must both exist and be equal:

$$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$$

If the function approaches different values from the left and right, the limit does not exist, and you have a jump discontinuity.

Condition 3: $\lim_{x \to a} f(x) = f(a)$. Even if the limit exists and the function is defined, they might not match. Imagine a function whose graph is a straight line, except at one point where someone has “moved” the point off the line. The limit would be where the point “should” be, but the actual function value is somewhere else.

Types of Discontinuities

When a function fails to be continuous at a point, understanding how it fails helps you understand the function better. There are three main types of discontinuities.

Removable Discontinuity (Hole)

A removable discontinuity occurs when the limit exists, but either $f(a)$ is undefined or $f(a)$ does not equal the limit. Graphically, this appears as a “hole” in the curve.

Consider $f(x) = \frac{x^2 - 1}{x - 1}$. For $x \neq 1$, we can simplify:

$$f(x) = \frac{(x-1)(x+1)}{x-1} = x + 1$$

So the graph of $f$ looks exactly like the line $y = x + 1$, except there is a hole at $x = 1$ (since the original function is not defined there). The limit as $x$ approaches 1 is 2, but $f(1)$ does not exist. This discontinuity is “removable” because we could define $f(1) = 2$ and the function would become continuous.

Jump Discontinuity

A jump discontinuity occurs when the left-hand limit and right-hand limit both exist but are not equal. The function “jumps” from one value to another.

A classic example is the floor function $\lfloor x \rfloor$, which rounds down to the nearest integer. At every integer, the function jumps. As $x$ approaches 2 from the left, $\lfloor x \rfloor$ stays at 1. As $x$ approaches 2 from the right, $\lfloor x \rfloor$ is already at 2. The function jumps from 1 to 2 at $x = 2$.

Infinite Discontinuity (Asymptotic Discontinuity)

An infinite discontinuity occurs when the function “blows up” to infinity (or negative infinity) as $x$ approaches $a$. This typically happens at vertical asymptotes.

The function $f(x) = \frac{1}{x}$ has an infinite discontinuity at $x = 0$. As $x$ approaches 0 from the right, $f(x)$ shoots off toward positive infinity. As $x$ approaches 0 from the left, $f(x)$ plunges toward negative infinity. The limit does not exist (not even as infinity, since the two sides disagree), and the function is wildly discontinuous at 0.

Continuity on Intervals

So far we have focused on continuity at a single point. But we often want to talk about functions being continuous over an entire interval.

A function is continuous on an open interval $(a, b)$ if it is continuous at every point in that interval.

For a closed interval $[a, b]$, we need to be slightly more careful at the endpoints. A function is continuous on $[a, b]$ if:

  • It is continuous at every point in $(a, b)$
  • $\lim_{x \to a^+} f(x) = f(a)$ (continuous from the right at $a$)
  • $\lim_{x \to b^-} f(x) = f(b)$ (continuous from the left at $b$)

We only check one-sided limits at the endpoints because we cannot approach from outside the interval.

A function is continuous everywhere (or simply “continuous”) if it is continuous at every point in its domain. Polynomials, sine and cosine, and exponential functions are all continuous everywhere they are defined.

Properties of Continuous Functions

Continuous functions play nicely with arithmetic operations. If $f$ and $g$ are both continuous at $x = a$, then:

  • $f + g$ is continuous at $a$
  • $f - g$ is continuous at $a$
  • $f \cdot g$ is continuous at $a$
  • $\frac{f}{g}$ is continuous at $a$, provided $g(a) \neq 0$
  • $f \circ g$ (the composition of $f$ and $g$) is continuous, under appropriate conditions

These properties explain why polynomials are continuous everywhere: they are built from the continuous function $f(x) = x$ using addition, subtraction, and multiplication. Rational functions (ratios of polynomials) are continuous everywhere except where the denominator is zero.

The Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is one of the most important results about continuous functions. It captures mathematically the intuition that a continuous function cannot “skip” values.

Intermediate Value Theorem: If $f$ is continuous on the closed interval $[a, b]$, and $N$ is any number between $f(a)$ and $f(b)$, then there exists at least one number $c$ in $(a, b)$ such that $f(c) = N$.

In plain language: if a continuous function starts at one value and ends at another, it must hit every value in between at least once.

This theorem has a remarkable application: it can prove that equations have solutions without actually solving them. If you can show that a continuous function is negative at one point and positive at another, the IVT guarantees that the function equals zero somewhere in between. That zero is a solution to the equation $f(x) = 0$.

The IVT also explains many everyday observations. If you start at sea level and climb a mountain, you pass through every altitude in between. You cannot go from 0 feet to 10,000 feet without passing through 5,000 feet. Your altitude as a function of time is continuous.

Notation and Terminology

Term Meaning Example
Continuous at $a$ $\lim_{x \to a} f(x) = f(a)$ Polynomials are continuous everywhere
Removable discontinuity Limit exists but $\neq f(a)$ or $f(a)$ undefined Hole in graph
Jump discontinuity Left and right limits differ Step functions, floor function
Infinite discontinuity Limit is $\pm\infty$ Vertical asymptote at $x = 0$ for $\frac{1}{x}$
Continuous on $(a,b)$ Continuous at every point in the interval $\sin(x)$ on $(0, \pi)$
Continuous on $[a,b]$ Continuous on $(a,b)$ plus one-sided continuity at endpoints $\sqrt{x}$ on $[0, 4]$
IVT Intermediate Value Theorem: continuous functions hit all intermediate values Used to prove existence of solutions

Examples

Example 1: Checking Continuity at a Point

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

Solution:

We need to verify the three conditions for continuity at $x = 3$.

Condition 1: Is $f(3)$ defined? $$f(3) = 3^2 = 9$$ Yes, $f(3) = 9$ is defined.

Condition 2: Does $\lim_{x \to 3} f(x)$ exist? $$\lim_{x \to 3} x^2 = 9$$ Yes, the limit exists and equals 9.

Condition 3: Does $\lim_{x \to 3} f(x) = f(3)$? $$\lim_{x \to 3} x^2 = 9 = f(3)$$ Yes, the limit equals the function value.

All three conditions are satisfied, so $f(x) = x^2$ is continuous at $x = 3$.

In fact, polynomials are continuous at every real number, so this result is not surprising. But it is good practice to verify using the definition.

Example 2: Identifying Types of Discontinuities

Classify the discontinuity at $x = 2$ for each function:

(a) $f(x) = \frac{x^2 - 4}{x - 2}$

(b) $g(x) = \begin{cases} 1 & \text{if } x < 2 \ 3 & \text{if } x \geq 2 \end{cases}$

(c) $h(x) = \frac{1}{(x-2)^2}$

Solution:

(a) We can factor the numerator: $\frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x + 2$ for $x \neq 2$.

The limit as $x \to 2$ is $\lim_{x \to 2} (x + 2) = 4$.

But $f(2)$ is undefined (division by zero).

The limit exists, but the function is not defined at the point. This is a removable discontinuity (a hole at $(2, 4)$).

(b) The left-hand limit: $\lim_{x \to 2^-} g(x) = 1$

The right-hand limit: $\lim_{x \to 2^+} g(x) = 3$

The one-sided limits exist but are not equal. This is a jump discontinuity.

(c) As $x \to 2$, the denominator $(x-2)^2$ approaches 0 through positive values, so: $$\lim_{x \to 2} \frac{1}{(x-2)^2} = +\infty$$

The function “blows up” to infinity. This is an infinite discontinuity (vertical asymptote at $x = 2$).

Example 3: Making a Piecewise Function Continuous

Find the value of $k$ that makes the following function continuous everywhere:

$$f(x) = \begin{cases} x^2 + 1 & \text{if } x < 2 \ kx - 3 & \text{if } x \geq 2 \end{cases}$$

Solution:

The function $x^2 + 1$ is a polynomial, so it is continuous for all $x < 2$.

The function $kx - 3$ is also a polynomial (linear), so it is continuous for all $x > 2$.

The only potential problem is at $x = 2$, where the two pieces meet. For continuity at $x = 2$, we need:

$$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$$

Left-hand limit (using the top formula): $$\lim_{x \to 2^-} (x^2 + 1) = 2^2 + 1 = 5$$

Right-hand limit (using the bottom formula): $$\lim_{x \to 2^+} (kx - 3) = k(2) - 3 = 2k - 3$$

Function value at $x = 2$ (using the bottom formula, since $x \geq 2$): $$f(2) = k(2) - 3 = 2k - 3$$

For continuity, we need the left-hand limit to equal the function value: $$5 = 2k - 3$$ $$8 = 2k$$ $$k = 4$$

Verification: With $k = 4$, the function becomes: $$f(x) = \begin{cases} x^2 + 1 & \text{if } x < 2 \ 4x - 3 & \text{if } x \geq 2 \end{cases}$$

At $x = 2$: Left limit $= 5$, right limit $= 4(2) - 3 = 5$, and $f(2) = 5$.

All three are equal, so $k = 4$ makes the function continuous everywhere.

Example 4: Using the Intermediate Value Theorem

Show that the equation $x^3 - x - 1 = 0$ has a solution in the interval $[1, 2]$.

Solution:

Let $f(x) = x^3 - x - 1$.

Step 1: Verify that $f$ is continuous on $[1, 2]$.

Since $f$ is a polynomial, it is continuous everywhere, including on $[1, 2]$.

Step 2: Evaluate $f$ at the endpoints.

$$f(1) = 1^3 - 1 - 1 = 1 - 1 - 1 = -1$$ $$f(2) = 2^3 - 2 - 1 = 8 - 2 - 1 = 5$$

Step 3: Apply the Intermediate Value Theorem.

We have $f(1) = -1 < 0$ and $f(2) = 5 > 0$.

Since $f$ is continuous on $[1, 2]$ and $0$ is between $f(1) = -1$ and $f(2) = 5$, the IVT guarantees there exists at least one value $c$ in $(1, 2)$ such that $f(c) = 0$.

Conclusion: The equation $x^3 - x - 1 = 0$ has at least one solution in the interval $(1, 2)$.

Note: The IVT tells us a solution exists but does not tell us what it is. The actual solution is approximately $c \approx 1.3247$, but finding it requires numerical methods. The power of the IVT is proving existence without computation.

Example 5: Continuity of $\frac{\sin x}{x}$ with a Defined Value at Zero

Consider the function:

$$f(x) = \begin{cases} \frac{\sin x}{x} & \text{if } x \neq 0 \ 1 & \text{if } x = 0 \end{cases}$$

Is $f$ continuous at $x = 0$?

Solution:

We need to check the three conditions for continuity at $x = 0$.

Condition 1: Is $f(0)$ defined?

Yes, by the piecewise definition, $f(0) = 1$.

Condition 2: Does $\lim_{x \to 0} f(x)$ exist?

For $x \neq 0$, $f(x) = \frac{\sin x}{x}$. This is one of the most important limits in calculus:

$$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

This limit can be proven using the squeeze theorem (comparing $\frac{\sin x}{x}$ to functions that bound it from above and below as $x \to 0$), or by geometric arguments involving the unit circle.

So yes, $\lim_{x \to 0} f(x) = 1$.

Condition 3: Does $\lim_{x \to 0} f(x) = f(0)$?

$$\lim_{x \to 0} f(x) = 1 = f(0)$$

Yes, the limit equals the function value.

Conclusion: All three conditions are satisfied, so $f$ is continuous at $x = 0$.

This example shows the power of the piecewise definition. The expression $\frac{\sin x}{x}$ is not defined at $x = 0$, but by choosing $f(0) = 1$ (the value of the limit), we “fill in the hole” and create a function that is continuous everywhere. This is an example of removing a removable discontinuity.

The function $f(x) = \frac{\sin x}{x}$ (with $f(0) = 1$) is called the sinc function and appears frequently in signal processing and Fourier analysis.

Example 6: A Subtle Continuity Analysis

Determine where the function $f(x) = \frac{x^2 - 5x + 6}{x^2 - 4}$ is continuous.

Solution:

First, let us factor both numerator and denominator.

Numerator: $x^2 - 5x + 6 = (x - 2)(x - 3)$

Denominator: $x^2 - 4 = (x - 2)(x + 2)$

So: $$f(x) = \frac{(x-2)(x-3)}{(x-2)(x+2)}$$

The function is undefined where the denominator is zero, which occurs at $x = 2$ and $x = -2$.

At $x = -2$:

Both the numerator and denominator approach 0, but let us check more carefully.

$$\lim_{x \to -2} \frac{(x-2)(x-3)}{(x-2)(x+2)} = \lim_{x \to -2} \frac{x-3}{x+2}$$

For $x \neq 2$, we can cancel the $(x-2)$ factors: $$= \lim_{x \to -2} \frac{x-3}{x+2} = \frac{-2-3}{-2+2} = \frac{-5}{0}$$

This limit does not exist (the function approaches $\pm\infty$). There is an infinite discontinuity (vertical asymptote) at $x = -2$.

At $x = 2$:

For $x \neq 2$: $$f(x) = \frac{(x-2)(x-3)}{(x-2)(x+2)} = \frac{x-3}{x+2}$$

$$\lim_{x \to 2} \frac{x-3}{x+2} = \frac{2-3}{2+2} = \frac{-1}{4}$$

The limit exists and equals $-\frac{1}{4}$, but $f(2)$ is undefined. This is a removable discontinuity (a hole at the point $(2, -\frac{1}{4})$).

Conclusion: The function $f(x) = \frac{x^2 - 5x + 6}{x^2 - 4}$ is continuous on $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$.

At $x = -2$, there is an infinite discontinuity. At $x = 2$, there is a removable discontinuity.

Key Properties and Rules

Continuity of Basic Functions

The following functions are continuous on their entire domains:

  • Polynomials: $f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ (continuous for all real $x$)
  • Rational functions: $f(x) = \frac{p(x)}{q(x)}$ (continuous wherever $q(x) \neq 0$)
  • Root functions: $f(x) = \sqrt[n]{x}$ (continuous on appropriate domain)
  • Trigonometric functions: $\sin x$, $\cos x$ (continuous everywhere); $\tan x$, $\sec x$ (continuous except at vertical asymptotes)
  • Exponential functions: $f(x) = a^x$ (continuous for all real $x$)
  • Logarithmic functions: $f(x) = \log_a x$ (continuous for $x > 0$)

Arithmetic Combinations

If $f$ and $g$ are continuous at $x = a$, then:

  • $(f + g)(x)$ is continuous at $a$
  • $(f - g)(x)$ is continuous at $a$
  • $(cf)(x)$ is continuous at $a$ for any constant $c$
  • $(fg)(x)$ is continuous at $a$
  • $\left(\frac{f}{g}\right)(x)$ is continuous at $a$, provided $g(a) \neq 0$

Composition of Functions

If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then the composite function $f \circ g$ is continuous at $a$:

$$\lim_{x \to a} f(g(x)) = f\left(\lim_{x \to a} g(x)\right) = f(g(a))$$

This rule is incredibly useful. It tells us, for example, that $e^{\sin x}$ is continuous everywhere (since both $e^x$ and $\sin x$ are continuous everywhere).

The Intermediate Value Theorem (Formal Statement)

If $f$ is continuous on $[a, b]$ and $N$ is any number strictly between $f(a)$ and $f(b)$, then there exists at least one $c \in (a, b)$ such that $f(c) = N$.

Corollary (Root-Finding Version): If $f$ is continuous on $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then there exists at least one $c \in (a, b)$ such that $f(c) = 0$.

Real-World Applications

Pricing Tiers vs. Smooth Pricing

Consider how different companies price their services. Some use pricing tiers: if you use up to 100 units, you pay $10; if you use 101-200 units, you pay $20. At exactly 100 units, there is a jump. This pricing function is discontinuous. Other companies use smooth pricing where the cost increases continuously with usage. Neither approach is inherently better, but understanding continuity helps you model and analyze both.

Physical Quantities are Typically Continuous

Temperature, pressure, velocity, and most physical quantities vary continuously over time. The temperature in a room does not suddenly jump from 68 degrees to 74 degrees. It must pass through every temperature in between. This is why differential equations (which rely on continuity and differentiability) are so effective at modeling physical systems.

When physical quantities do appear to change discontinuously (like a light switch going from “off” to “on”), we are usually approximating a very rapid continuous change, or we are measuring a discrete quantity (like the number of photons).

The IVT and Crossing Sea Level

Suppose you start in a cave 100 feet below sea level and hike to a mountain peak 5,000 feet above sea level. The Intermediate Value Theorem guarantees that at some point during your hike, you were at exactly sea level (0 feet elevation). You cannot go from -100 feet to +5,000 feet without passing through 0. Your altitude as a function of position along your path is continuous.

This may seem obvious, but the IVT makes it mathematically rigorous. And the rigor matters when we apply the same reasoning to less intuitive situations, like proving that certain equations have solutions.

The Round-Trip Theorem

Here is a beautiful application of the IVT. Suppose you climb a mountain, starting at 6 AM and reaching the summit by 6 PM. The next day, you descend, again starting at 6 AM and reaching the bottom by 6 PM. The IVT guarantees that there is some time of day (say, 2:47 PM) when you were at exactly the same altitude on both days.

To see why, define $f(t) = (\text{altitude on day 1 at time } t) - (\text{altitude on day 2 at time } t)$.

At 6 AM: You are at the bottom on day 1 and at the top on day 2, so $f(\text{6 AM}) < 0$.

At 6 PM: You are at the top on day 1 and at the bottom on day 2, so $f(\text{6 PM}) > 0$.

Since $f$ is continuous (altitude varies continuously with time) and changes sign, the IVT says $f(t) = 0$ for some time $t$. At that moment, your altitude on day 1 equals your altitude on day 2.

Bisection Method for Finding Roots

The IVT is the theoretical foundation for the bisection method, a simple but powerful algorithm for finding where a function equals zero. If $f(a) < 0$ and $f(b) > 0$, the IVT says there is a root in $(a, b)$. Check the midpoint $c = \frac{a+b}{2}$. If $f(c) = 0$, you found the root. If $f(c) < 0$, the root is in $(c, b)$. If $f(c) > 0$, the root is in $(a, c)$. Repeat, and you will zero in on the root with arbitrary precision.

Self-Test Problems

Problem 1: Determine whether $f(x) = \frac{1}{x-4}$ is continuous at $x = 4$. Explain your answer.

Show Answer

No, $f(x) = \frac{1}{x-4}$ is not continuous at $x = 4$.

The function is not defined at $x = 4$ (division by zero), so the first condition for continuity fails immediately. Additionally, $\lim_{x \to 4} \frac{1}{x-4}$ does not exist (the function approaches $+\infty$ from the right and $-\infty$ from the left).

There is an infinite discontinuity (vertical asymptote) at $x = 4$.

Problem 2: Classify the discontinuity of $f(x) = \frac{x^2 - 9}{x - 3}$ at $x = 3$.

Show Answer

This is a removable discontinuity.

Factor the numerator: $\frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} = x + 3$ for $x \neq 3$.

The limit as $x \to 3$ is $\lim_{x \to 3}(x + 3) = 6$, which exists.

However, $f(3)$ is undefined (the original expression has $0$ in the denominator).

Since the limit exists but the function is undefined at the point, this is a removable discontinuity. There is a “hole” in the graph at the point $(3, 6)$.

Problem 3: Find all values of $k$ that make the following function continuous at $x = 1$: $$f(x) = \begin{cases} 3x + k & \text{if } x \leq 1 \ kx^2 + 2 & \text{if } x > 1 \end{cases}$$

Show Answer

For continuity at $x = 1$, the left-hand limit must equal the right-hand limit, which must equal $f(1)$.

Left-hand limit (and $f(1)$, since we use the top formula when $x \leq 1$): $$\lim_{x \to 1^-} (3x + k) = 3(1) + k = 3 + k$$

Right-hand limit: $$\lim_{x \to 1^+} (kx^2 + 2) = k(1)^2 + 2 = k + 2$$

For continuity: $3 + k = k + 2$

Solving: $3 + k = k + 2 \Rightarrow 3 = 2$

This is a contradiction. There is no value of $k$ that makes this function continuous at $x = 1$.

No matter what value of $k$ we choose, the function will have a jump discontinuity at $x = 1$ (the jump is always from $3 + k$ to $k + 2$, a jump of $-1$).

Problem 4: Use the Intermediate Value Theorem to show that $\cos x = x$ has a solution in the interval $[0, 1]$.

Show Answer

Let $f(x) = \cos x - x$.

If $f(c) = 0$, then $\cos c = c$, so we need to show $f$ has a root in $[0, 1]$.

Step 1: $f(x) = \cos x - x$ is continuous on $[0, 1]$ (it is the difference of two continuous functions).

Step 2: Evaluate at the endpoints. $$f(0) = \cos 0 - 0 = 1 - 0 = 1 > 0$$ $$f(1) = \cos 1 - 1 \approx 0.5403 - 1 = -0.4597 < 0$$

Step 3: Since $f$ is continuous on $[0, 1]$ and $f(0) > 0 > f(1)$, the IVT guarantees there exists $c \in (0, 1)$ such that $f(c) = 0$.

Therefore, the equation $\cos x = x$ has at least one solution in $(0, 1)$.

(The actual solution is approximately $x \approx 0.7391$, known as the Dottie number.)

Problem 5: Where is $f(x) = \sqrt{4 - x^2}$ continuous? Explain.

Show Answer

The function $f(x) = \sqrt{4 - x^2}$ is continuous on $[-2, 2]$.

For the square root to be defined, we need $4 - x^2 \geq 0$, which means $x^2 \leq 4$, so $-2 \leq x \leq 2$.

The domain of $f$ is $[-2, 2]$.

On the interior $(-2, 2)$, the function $4 - x^2$ is a polynomial (continuous everywhere), and it is positive on this interval. The square root of a positive continuous function is continuous, so $f$ is continuous on $(-2, 2)$.

At the endpoints:

  • $\lim_{x \to -2^+} \sqrt{4 - x^2} = \sqrt{0} = 0 = f(-2)$
  • $\lim_{x \to 2^-} \sqrt{4 - x^2} = \sqrt{0} = 0 = f(2)$

So $f$ is continuous from the right at $-2$ and from the left at $2$.

Therefore, $f$ is continuous on its entire domain $[-2, 2]$.

Geometrically, the graph of $f$ is the upper half of a circle with radius 2, centered at the origin. It is a smooth, unbroken curve from $(-2, 0)$ to $(2, 0)$.

Summary

  • A function is continuous at $x = a$ if three conditions hold: $f(a)$ is defined, $\lim_{x \to a} f(x)$ exists, and $\lim_{x \to a} f(x) = f(a)$.

  • The intuitive test for continuity is whether you can draw the graph “without lifting your pencil.” No jumps, no holes, no asymptotes.

  • Discontinuities come in three types: removable (a hole that could be filled), jump (the function leaps from one value to another), and infinite (the function blows up to infinity).

  • Polynomials are continuous everywhere. Rational functions are continuous wherever the denominator is nonzero. Most “elementary” functions (trig, exponential, logarithmic) are continuous on their natural domains.

  • Continuous functions are well-behaved under arithmetic: sums, differences, products, quotients (where defined), and compositions of continuous functions are continuous.

  • The Intermediate Value Theorem says that if $f$ is continuous on $[a, b]$ and $N$ is between $f(a)$ and $f(b)$, then $f(c) = N$ for some $c$ in $(a, b)$. This is the mathematical guarantee that continuous functions “hit all values in between.”

  • The IVT is a powerful tool for proving existence of solutions. If a continuous function changes sign on an interval, it must have a root somewhere in that interval.

  • Continuity connects the intuitive world of smooth, predictable change to the rigorous world of limits. It is the foundation for the deeper results of calculus that follow.