Continuity and the Intermediate Value Theorem

• State the three conditions that make a function continuous at a point, and use them to decide whether a given function is continuous there.
• Classify a discontinuity as removable, jump, infinite, or oscillating, and say which kind can be repaired by redefining a single value.
• Build continuous functions from known continuous pieces, and choose a parameter so that a piecewise function is continuous.
• State the Intermediate Value Theorem with all of its hypotheses, and show by example what goes wrong when each one is dropped.
• Explain why the theorem is the existence guarantee behind bisection search, and recognize inputs where a sign change does not guarantee a root.

The lesson guess-and-check-and-bisection-search solves equations that have no formula by halving an interval: keep a low and a high, test the midpoint, discard the half that cannot contain the answer. Its square-root loop works, and the lesson explains why it is fast. What it leaves open is a prior question: why is there an answer inside [low, high] at all? A program can halve an interval forever without ever being told that the thing it is hunting exists.

Calculus answers that question with the Intermediate Value Theorem. Roughly, a function whose graph has no breaks cannot go from negative to positive without passing through zero. The word "roughly" hides the whole difficulty, because "no breaks" has to become a precise, checkable property of $f$. That property is continuity, and it is the first place in this discipline where a limit is used to define a property of a function rather than to compute a number.

Continuity also matters beyond bisection. Nearly every theorem about derivatives and integrals in the lessons ahead begins with the words "suppose $f$ is continuous on $[a, b]$", and the failures are instructive: a discontinuity is where numerical methods, optimizers and simulations tend to misbehave, because a small change in the input produces a large change in the output.

Continuity at a point

A function $f$ is continuous at $a$ when all three of the following hold:

1. $f(a)$ is defined.
2. $\lim_{x \to a} f(x)$ exists.
3. $\lim_{x \to a} f(x) = f(a)$.

In words: the value the function approaches (limits-intuition-and-the-formal-definition) is the value it actually takes. Equivalent form: $f(a + h) \to f(a)$ as $h \to 0$, so a tiny change in the input produces a tiny change in the output.

A function is continuous on an interval if it is continuous at every point of it. At the endpoints of a closed interval $[a, b]$ only the one-sided limit from inside the interval is required.

Continuity is built up from pieces. Polynomials, $\sin x$, $\cos x$ and $e^x$ are continuous at every real number. Sums, differences, products and compositions of continuous functions are continuous, and a quotient $f/g$ is continuous wherever $g \neq 0$. These facts are the limit laws (limit-laws-and-indeterminate-forms) restated, which is why most continuity checks reduce to looking for a place where a denominator vanishes or a formula changes.

Kinds of discontinuity

When one of the three conditions fails, the way it fails has a name:

• Removable: $\dfrac{x^2 - 1}{x - 1}$ at $1$. The limit is $2$, but $f(1)$ is undefined.
• Jump: $\lfloor x \rfloor$ at $0$. The one-sided limits are $-1$ and $0$, so the limit does not exist.
• Infinite: $\dfrac{1}{x}$ at $0$. The function is unbounded near $a$.
• Oscillating: $\sin\!\left(\dfrac{1}{x}\right)$ at $0$. The values keep swinging between $-1$ and $1$.

Only the removable kind can be repaired, by assigning $f(a)$ the value of the limit. Defining $f(1) = 2$ turns $\frac{x^2-1}{x-1}$ into the continuous function $x + 1$.

A point where $f$ is not defined is a point where $f$ is not continuous, but that does not make $\frac{1}{x}$ a "broken" function: it is continuous at every point of its domain, which simply has a gap at $0$. This distinction returns in the theorem below.

The Intermediate Value Theorem

Theorem. Let $f$ be continuous on the closed interval $[a, b]$, and let $N$ be any number strictly between $f(a)$ and $f(b)$. Then there is at least one $c$ in $(a, b)$ with $f(c) = N$.

The most used special case is $N = 0$: if $f$ is continuous on $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then $f$ has a root in $(a, b)$.

Each hypothesis earns its place, and the fastest way to see it is to remove it.

• Drop continuity. Let $f(x) = -1$ for $x < 0$ and $f(x) = 1$ for $x \geq 0$, on $[-1, 1]$. The endpoint values are $-1$ and $1$, yet $f$ never equals $0$, or any other number in between.
• Drop the interval. $f(x) = \frac{1}{x}$ has $f(-1) = -1$ and $f(1) = 1$, and is continuous at every point where it is defined, yet it has no root. The interval $[-1, 1]$ is not inside the domain, because $0$ is missing.
• Reverse the implication. The theorem does not say that a root forces a sign change. $f(x) = x^2$ has a root at $0$ on $[-1, 1]$ although $f(-1) = f(1) = 1$.
• Change the number system. The theorem relies on the real numbers having no gaps. On the rationals, $f(x) = x^2 - 2$ is continuous, $f(0) = -2$ and $f(2) = 2$, and there is still no rational root.

The conclusion is also weak in two ways worth stating. It says "at least one", not exactly one: $x^3 - x$ has three roots on $[-2, 2]$. And it says nothing about where $c$ is. It is a statement of existence, and finding $c$ is a separate job.

From the theorem to bisection

Suppose $f$ is continuous on $[a, b]$ and $f(a)$, $f(b)$ have opposite signs. Let $m = \frac{a + b}{2}$. If $f(m) = 0$ the search is over. Otherwise $f(m)$ has the sign of exactly one endpoint, so one of the halves $[a, m]$ and $[m, b]$ again has endpoints of opposite signs. The theorem, applied to that half, guarantees a root inside it. Repeating the step gives intervals of width $\frac{b - a}{2^n}$ after $n$ steps, each containing a root.

This is the loop of guess-and-check-and-bisection-search. Its square-root search for $x > 0$ is bisection on $f(t) = t^2 - x$ over $[0, \max(1, x)]$. At the left end $f(0) = -x < 0$, at the right end $f(\text{high}) \geq 0$ (for $x \geq 1$ it is $x^2 - x$; for $x < 1$ it is $1 - x$). The test guess ** 2 < x is the sign of $f(\text{guess})$. A polynomial is continuous, so the theorem is what makes the invariant "a root is between low and high" true at every step. That lesson states the precondition as an ordered space where each check reveals a direction. The theorem gives a weaker and more general one: a sign change on an interval where $f$ is continuous, with no need for $f$ to be monotone.

Example 1: choosing a parameter for continuity

Find $a$ so that

is continuous everywhere.

Each piece is a polynomial, hence continuous on its own region, so the only risk is at $x = 2$. The left piece gives $f(2) = 2a + 1$, which is also the limit from the left. The limit from the right is $2^2 - a = 4 - a$. Condition 3 asks for $2a + 1 = 4 - a$, so $a = 1$.

Check with $a = 1$: both sides give $3$, so the limit exists and equals $f(2) = 3$. With any other $a$ the one-sided limits differ and $f$ has a jump at $2$.

Example 2: solving $\cos x = x$ by bisection

Let $f(x) = \cos x - x$ on $[0, 1]$. It is continuous (a difference of continuous functions), $f(0) = 1 > 0$ and $f(1) \approx -0.4597 < 0$. The theorem guarantees a root in $(0, 1)$. Bisection, keeping the half whose endpoints still have opposite signs:

text
step     a         b         m       f(m)      keep
1     0.000000  1.000000  0.500000  +0.377583  right
2     0.500000  1.000000  0.750000  -0.018311  left
3     0.500000  0.750000  0.625000  +0.185963  right
4     0.625000  0.750000  0.687500  +0.085335  right
5     0.687500  0.750000  0.718750  +0.033879  right
6     0.718750  0.750000  0.734375  +0.007875  right
7     0.734375  0.750000  0.742188  -0.005196  left
8     0.734375  0.742188  0.738281  +0.001345  right

After eight steps the root lies in $[0.738281, 0.742188]$, an interval of width $2^{-8} \approx 0.0039$. The true root is $0.7390851\ldots$, inside it, as the theorem requires. To pin the root down to width $10^{-6}$ starting from width $1$ takes the smallest $n$ with $2^{-n} \leq 10^{-6}$, which is $n = 20$.

This function has no closed-form solution, which is exactly the situation in which the theorem and the algorithm earn their keep.

Example 3: a sign change that does not contain a root

Take $f(x) = \frac{1}{x}$ on $[-1, 2]$. The endpoint values are $-1$ and $0.5$, opposite signs, so the bisection loop starts without complaint. The midpoint sequence squeezes toward $0$:

text
step   interval after the step       f(m)
1      [-1.0, 0.5]                    +2
5      [-0.0625, 0.03125]             +32
10     [-0.00098, 0.00195]            -1024
20     [-9.5e-07, 1.9e-06]            -1048576

The interval shrinks as promised, but $\lvert f(m) \rvert$ doubles at every step instead of shrinking. The loop is converging to the discontinuity, not to a root. A practical guard is to check, at the end, that $\lvert f \rvert$ at the final point is actually small. A sign change tells you a root exists only when the continuity hypothesis has been checked separately.

• "Continuous means you can draw the graph without lifting the pen." That picture is fair for a function on an interval, and it is how the theorem is usually remembered. It misleads on domains with gaps: $\frac{1}{x}$ is continuous at every point where it is defined, even though its graph comes in two pieces.
• "The theorem finds the root, or says it is unique." It only asserts that at least one exists in the open interval. Where, and how many, are separate questions. Bisection on $x^3 - x$ over $[-2, 1.5]$ converges to $-1$ and ignores the roots at $0$ and $1$.
• "If there is no sign change, there is no root." $x^2$ on $[-1, 1]$ touches zero without changing sign, so bisection started there has nothing to work with. The theorem gives a sufficient condition, not a necessary one.
• "Any bracket with opposite signs makes bisection safe." Example 3 is the counter-example. Continuity on the whole bracket is part of the hypothesis, and bisection cannot check it.
• "On a computer, bisection halves forever." Real numbers can be halved indefinitely; doubles cannot. Applied to $\cos x - x$ on $[0, 1]$ in double precision, a loop with no zero test halves the interval 53 times before the midpoint equals an endpoint, leaving a width of about $1.1 \times 10^{-16}$. With the zero test of the algorithm above, the run stops one step earlier, at step 52, where cos(m) - m evaluates to exactly 0.0, which is a rounding coincidence and not a proof that $m$ is the exact root. A loop that stops when the midpoint repeats an endpoint always ends. A loop that stops on interval width ends only if the tolerance is no smaller than the spacing of doubles near the root, about $1.1 \times 10^{-16}$ here, so while b - a > 1e-20 never exits. A loop that demands $\lvert f(m) \rvert$ below a very small tolerance may never get there, as the bisection lesson warns.

A function is continuous at $a$ when $f(a)$ is defined, the limit at $a$ exists, and the two agree. Failures come in four kinds: removable, jump, infinite and oscillating, and only the first can be repaired by redefining one value. The Intermediate Value Theorem says that if $f$ is continuous on $[a, b]$ then $f$ takes every value between $f(a)$ and $f(b)$ at some point of the interval; with opposite signs at the ends it guarantees a root. Each hypothesis is needed: dropping continuity, using an interval with a gap, or leaving the real numbers breaks the conclusion, and the converse is false. Bisection is the theorem turned into an algorithm: each step keeps the half with a sign change, so the root stays trapped while the width falls like $2^{-n}$. The theorem guarantees existence, not uniqueness or location, and bisection cannot detect a discontinuity on its own. The next lesson, the-derivative-as-a-rate-of-change, uses the same limit machinery to ask how fast $f$ changes at a point.