2.4 Continuity

Big idea

So far the value of a function at a point and the limit of the function at that point have been kept deliberately apart. Continuity is the name for the situation where they finally coincide. A function is continuous at a point when it is defined there, has a limit there, and the two numbers match. Those three requirements are not redundant; each can fail on its own, and the failures look different on a graph.

The reason to isolate this condition is that continuity is what makes substitution legal. The limit laws already gave you substitution for polynomials and for rational functions away from their zeros in the denominator. Continuity generalizes the pattern and names it: for a continuous function, computing a limit is evaluating the function, and the whole apparatus of tables and cancellation can be skipped. Since sums, products, quotients, compositions, roots, trigonometric functions, exponentials and logarithms are all continuous where they are defined, almost every function you write down is continuous almost everywhere, and the exceptions are worth listing precisely.

Continuity also carries real content of its own, not just convenience. A continuous function on a closed interval cannot get from one value to another without passing through everything in between. That is the intermediate value theorem, and it is the first result in the course that asserts something exists without producing it. It is how you know an equation has a solution before you have any idea what the solution is, and it turns the informal image of an unbroken curve into a tool.

Piecewise functions are where all of this is tested. A formula that switches at a boundary is continuous away from the boundary for free, so the work happens at the seam: compute the limit from each side, compute the value, and compare. When a constant is left unspecified, that comparison becomes an equation to solve, which is the standard way these problems are posed.

Decoder

A function is continuous at a point when it is defined there, its limit there exists, and the limit equals the value.

Three conditions, three failure modes. If the function is undefined at the point you get a hole or an asymptote. If the limit fails to exist you get a jump or a blow-up. If both exist but disagree you get a hole with a stray point plotted somewhere else. Naming which condition failed is usually the whole answer to a classification question.

The compressed form $\lim_{x\to a} f(x) = f(a)$ silently asserts all three at once, since writing $f(a)$ presumes it exists and writing an equation presumes the left side exists too. It is worth unpacking every time, because on a piecewise function the three checks are three separate computations.

The classic mistake is to treat continuity as a property of a formula rather than of a function at a point. The function $1/x$ is continuous at every point of its domain; it is not discontinuous at $x = 0$, because it is not anything at $x = 0$. When a question asks where a function is continuous, the honest answer is a set of points, and the domain is where that answer starts.

Definitions and results

Continuity at a point. $f$ is continuous at $a$ when $f(a)$ is defined, $\lim_{x\to a} f(x)$ exists, and

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

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

One-sided continuity. $f$ is continuous from the left at $a$ when $\lim_{x\to a^-} f(x) = f(a)$, and from the right when $\lim_{x\to a^+} f(x) = f(a)$. Continuity at $a$ is exactly both at once.

Continuity on an interval. $f$ is continuous on an open interval when it is continuous at every point of it. On a closed interval $[a,b]$ the endpoints are only asked for the one-sided condition that points into the interval.

Removable discontinuity. The limit exists but the function is undefined there or defined with a different value. Reassigning that single value makes the function continuous, which is what “removable” means. Graphically: a hole.

Jump discontinuity. Both one-sided limits exist and differ. No reassignment of a single value can fix it, since the two sides disagree about the destination. Graphically: a step.

Infinite discontinuity. At least one one-sided limit is $\infty$ or $-\infty$. The line through the point is a vertical asymptote.

Which functions are continuous. Polynomials are continuous everywhere. Rational functions, root functions, trigonometric functions, exponentials and logarithms are continuous at every point of their domains. Sums, differences, products and quotients of continuous functions are continuous wherever the result is defined, and a composition $f(g(x))$ is continuous at $a$ when $g$ is continuous at $a$ and $f$ is continuous at $g(a)$.

Limits through a continuous outer function. If $f$ is continuous at $L$ and $\lim_{x\to a} g(x) = L$, then $\lim_{x\to a} f\big(g(x)\big) = f(L)$. This is what lets you move a limit inside a square root or a sine.

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 at least one $c$ in $[a,b]$ with $f(c) = N$. The usual application takes $N = 0$: a sign change on a closed interval forces a root inside it. The theorem promises existence and says nothing about how many or where.

Worked examples

A hole that can be patched

Let $f(x) = \dfrac{x^2 - x - 6}{x - 3}$. The denominator vanishes at $x = 3$, so $f(3)$ is undefined and the first condition fails. The numerator factors as $(x-3)(x+2)$, so for $x \ne 3$ the function equals $x + 2$ and

$$ \lim_{x\to 3} f(x) = 5 $$

The limit exists, so the discontinuity is removable. Defining $f(3) = 5$ produces a function continuous on the whole real line. Any other choice of value, say $f(3) = 0$, leaves a discontinuity of the same removable type, now of the second kind: limit and value both exist but disagree.

Choosing a constant to close a seam

Find the value of $a$ making

$$ f(x) = \begin{cases} x^2 + a, & x < 3 \\ 2x - 1, & x \ge 3 \end{cases} $$

continuous everywhere.

Away from $x = 3$ each branch is a polynomial, so continuity there is automatic. At the seam, the left-hand limit uses the first branch and equals $9 + a$. The right-hand limit uses the second branch and equals $5$, which is also the value $f(3) = 2(3) - 1 = 5$. Continuity requires all three to agree:

$$ 9 + a = 5 \implies a = -4 $$

Check by rebuilding. With $a = -4$ the left branch is $x^2 - 4$, whose value approaches $9 - 4 = 5$ as $x$ rises to $3$, matching $f(3) = 5$. The seam closes.

Sorting the discontinuities of a rational function

Let $g(x) = \dfrac{x^2 - 4}{x^2 - 3x + 2}$. Factor both parts:

$$ g(x) = \frac{(x-2)(x+2)}{(x-1)(x-2)} $$

The domain excludes $x = 1$ and $x = 2$, and those are the only candidates. At $x = 2$ the common factor cancels, leaving $\frac{x+2}{x-1}$ for $x \ne 2$, whose limit is $\frac{4}{1} = 4$. So $x = 2$ is a removable discontinuity.

At $x = 1$ nothing cancels: the numerator tends to $-3$ while the denominator tends to $0$, changing sign as $x$ crosses $1$. The one-sided limits are $\infty$ and $-\infty$, so $x = 1$ is an infinite discontinuity and $x = 1$ is a vertical asymptote. Everywhere else $g$ is a quotient of polynomials with a nonzero denominator, hence continuous.

Trapping a root

Show that $x^3 + x - 5 = 0$ has a solution between $1$ and $2$, and narrow it.

Let $f(x) = x^3 + x - 5$, a polynomial, so it is continuous on $[1,2]$ and the theorem applies. Evaluate the endpoints:

$$ f(1) = 1 + 1 - 5 = -3, \qquad f(2) = 8 + 2 - 5 = 5 $$

Zero lies between $-3$ and $5$, so some $c$ in $(1,2)$ has $f(c) = 0$.

To narrow it, test the midpoint: $f(1.5) = 3.375 + 1.5 - 5 = -0.125$, still negative, so the sign change now lives in $(1.5, 2)$. Testing again, $f(1.6) = 4.096 + 1.6 - 5 = 0.696$ is positive, so the root is in $(1.5, 1.6)$. Repeating the halving is a complete root-finding method, and every step of it rests on continuity.

Practice

Start with the three-part test on ordinary functions. You decide whether a function is continuous at a given point and, when it is not, name the type of discontinuity.

Then the seams. These drills give a piecewise definition, sometimes with an unknown constant, and ask you to check or enforce continuity at the boundary.

Last, existence arguments. You verify the hypotheses on a closed interval and conclude that a value or a root must occur inside it.

Quiz

Six items on the continuity test, classifying discontinuities, continuity of piecewise functions, and the intermediate value theorem.