Continuity

In many cases, you can compute $\displaystyle \lim_{x\to a} f(x)$ by plugging a in for x:

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

For example,

$$\lim_{x\to 3} (2 x^3 - 5 x + 1) = 2 \cdot 3^3 - 5 \cdot 3 + 1 = 40.$$

This situation arises often enough that it has a name.

Definition. A function $f(x)$ is continuous at a if

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

This definition really comprises three things, each of which you need to check to show that f is continuous at a:

1. $f(a)$ is defined.

2. $\displaystyle \lim_{x \to a} f(x)$ is defined.

3. The two are equal: $\displaystyle \lim_{x \to a} f(x) = f(a)$ .

What does this mean geometrically? Here are the three criteria above in pictorial language:

1. "$f(a)$ is defined" means there's a point on the graph at a.

2. "$\displaystyle \lim_{x \to a} f(x)$ is defined" means the graph approaches a single numerical value as you get close to a.

3. "$\displaystyle \lim_{x \to a} f(x) = f(a)$ " means that the value you're approaching is the value that f actually takes on --- there are "no surprises".

The first criterion means that there can't be a hole or gap in the graph. This also rules out vertical asymptotes. Here are some pictures of these kinds of discontinuities:

$$\hbox{\epsfysize=2.2in \epsffile{continuity1.eps}}$$

The second criterion means that the graph can't "jump" at a. This is a jump discontinuity:

$$\hbox{\epsfysize=1.1in \epsffile{continuity2.eps}}$$

A jump discontinuity occurs when the left and right-hand limits aren't equal.

The third criterion means that the graph is "filled in" at $x = a$ as you'd expect. You don't get close to a expecting one value and then find that $f(a)$ is something different, as you do below:

$$\hbox{\epsfysize=1.1in \epsffile{continuity3.eps}}$$

This is called a removable discontinuity because you could make the function continuous by filling in the hole. In terms of limits, it means that $\displaystyle \lim_{x\to a} f(x)$ exists, but $\displaystyle \lim_{x\to a} f(x) \ne f(a)$ .

Here are some classes of continuous functions:

(a) A polynomial $p(x)$ is continuous for all x.

(b) $|x|$ is continuous for all x.

(c) Trigonometric functions are continuous wherever they are defined.

(d) $e^x$ is continuous for all x and $\ln x$ is continuous for $x > 0$ .

(e) $\root n \of x$ is continuous for all x for which it's defined.

The statement about polynomials, for example, follows from a property of limits. If $p(x)$ is a polynomial, I showed that

$$\lim_{x \to c} p(x) = p(c).$$

This is exactly what it means for $p(x)$ to be continuous at c.

For example, $f(x) = 10 x^{100} - 15 x + 41$ is continuous for all x, since it's a polynomial.

The statement for $|x|$ is a fairly easy limit proof, but I'll omit it. And the statements for trig functions, $e^x$ , and $\ln x$ depend on careful definitions of these functions; I'll discuss some of this later. As an example of the statement about trig functions, $\tan x$ is continuous for all x except odd multiples of $\dfrac{\pi}{2}$ . ($\tan x$ is undefined at odd multiples of $\dfrac{\pi}{2}$ .)

The limit proof for $\root n \of x$ is not too hard, but once you have results on $e^x$ and $\ln x$ , you can use the fact that

$$\root n \of x = x^{1/n} = e^{(\ln x)/n}.$$

You have to ensure that the root you're taking is defined at the point. For example, $\sqrt{x}$ is continuous for $x \ge 0$ . Remember that $\sqrt{x}$ is undefined for $x < 0$ .

Example. What kind of discontinuity does $f(x) = \dfrac{x^2 - 1}{x^2 - x - 2}$ have at $x = -1$ ?

Note that

$$\lim_{x\to -1} \dfrac{x^2 - 1}{x^2 - x - 2} = \lim_{x\to -1} \dfrac{(x - 1)(x + 1)}{(x - 2)(x + 1)} = \lim_{x\to -1} \dfrac{x - 1}{x - 2} = \dfrac{2}{3}.$$

However, $f(-1)$ is undefined. Therefore, there is a removable discontinuity at $x = -1$ . I could make the function continuous at $x = -1$ by defining $f(-1) = \dfrac{2}{3}$ .

Example. Let

$$f(x) = \cases{ x^2 + 3 & if $x > 2$ \cr 4 x & if $x \le 2$ \cr}.$$

What kind of discontinuity does $f(x)$ have at $x = 2$ ?

Note that

$$\lim_{x\to 2^+} f(x) = \lim_{x\to 2^+} (x^2 + 3) = 7 \quad\hbox{and}\quad \lim_{x\to 2^-} f(x) = \lim_{x\to 2^-} 4 x = 8.$$

Thus, the left and right-hand limits aren't equal. Therefore, there is a jump discontinuity at $x = 2$ .

You can also get continuous functions by combining continuous functions in various ways.

Theorem. (a) If f and g are continuous at $x = c$ , so is their sum $f + g$ .

(b) If f and g are continuous at $x = c$ , so is their difference $f - g$ .

(c) If f and g are continuous at $x = c$ , so is their product $f g$ .

(d) If f and g are continuous at $x = c$ , and if $g(c) \ne 0$ , then the quotient $\dfrac{f}{g}$ is continuous at $x = c$ .

(e) If f is continuous at $g(c)$ and if g is continuous at $x = c$ , then the composite $f\circ g$ is continuous at $x = c$ .

Proof. The proofs are fairly easy consequences of our theorems on limits. I'll prove (c) by way of example.

Suppose f and g are continuous at $x = c$ . This means that

$$\lim_{x \to c} f(x) = f(c) \quad\hbox{and}\quad \lim_{x \to c} g(x) = g(c).$$

Multiplying the left sides and the right sides, I get

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

By the rule for the limit of a product,

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

Therefore,

$$\lim_{x \to c} f(x) g(x) = f(c) g(c).$$

This is what it means for $f(x) g(x)$ to be continuous at $x = c$ .

Here are some illustrations of these rules.

Since $e^x$ and $x^3$ are continuous for all x, their sum $e^x + x^3$ and their product $x^3 e^x$ are continuous for all x.

The quotient $\dfrac{e^x}{x^3}$ is continuous for all x except $x = 0$ (where the quotient is undefined).

Composition is an important method for constructing continuous functions. For example, $f(x) = \sin x$ is continuous for all x. The polynomial $g(x) = x^4 - 7 x^2 + x + 1$ is also continuous for all x. The composite is

$$f(g(x)) = \sin (x^4 - 7 x^2 + x + 1).$$

It is continuous for all x.

Example. Let

$$f(x) = \cases{ 1 & if $x < 0$ \cr x^2 & if $0 \le x < 1$ \cr \sin \dfrac{\pi x}{2} & if $x \ge 1$}.$$

For what values of x is f continuous?

The function is continuous except possibly at the "break points" between the three pieces. I must check the points $x = 0$ and $x = 1$ separately.

$$\hbox{\epsfysize=1.75in \epsffile{continuity4.eps}}$$

At $x = 0$ ,

$$\lim_{x \to 0-} f(x) = 1 \quad\hbox{but}\quad \lim_{x \to 0+} f(x) = 0.$$

Since the left- and right-hand limits do not agree,

$$\lim_{x \to 0} f(x) \hbox{ is undefined}.$$

Hence, f is not continuous at $x = 0$ .

At $x = 1$ ,

$$\lim_{x \to 1-} f(x) = \lim_{x \to 1-} x^2 = 1, \quad\hbox{and}\quad \lim_{x \to 1+} f(x) = \lim_{x \to 1+} \sin \dfrac{\pi x}{2} = 1.$$

The left- and right-hand limits agree, so

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

Now $f(1) = 1$ , so

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

Therefore, f is continuous at $x = 1$ .

Conclusion: f is continuous for all x except $x = 0$ .

Example. Let

$$f(x) = \dfrac{x^2 - 2 x - 3}{x^2 - 1}.$$

For what values of x is f continuous?

f is a quotient of two polynomials, and polynomials are continuous for all x. Hence, $f(x)$ is continuous at all points except those which make the bottom equal to 0.

Write f as

$$f(x) = \dfrac{(x - 3)(x + 1)}{(x - 1)(x + 1)}.$$

Hence, f is continuous for all x except $x = 1$ and $x = -1$ . (Note that you can't cancel the $x + 1$ -terms before seeing where f is undefined.)

However, the discontinuity at $x = -1$ is a removable discontinuity:

$$\lim_{x \to -1} f(x) = \lim_{x \to -1} \dfrac{x^2 - 2 x - 3}{x^2 - 1} = \lim_{x \to -1} \dfrac{(x - 3)(x + 1)}{(x - 1)(x + 1)} = \lim_{x \to -1} \dfrac{x - 3}{x - 1} = \dfrac{-4}{-2} = 2.$$

$f(-1)$ is undefined, but if I defined $f(-1) = 2$ , then the new f would be continuous at $x = -1$ .

On the other hand, the discontinuity at $x = 1$ is a vertical asymptote; no matter how I define $f(1)$ , the function will still be discontinuous at $x = 1$ .

Continuous functions possess the intermediate value property. Roughly put, it says that a if continuous function goes from one value to another, it doesn't skip any values in between. This corresponds to the geometric intuition that the graph of a continuous function doesn't have any gaps, jumps, or holes. Here is the precise statement.

Theorem. ( Intermediate Value Theorem) Let $f(x)$ be a continuous function on the interval $a \le x \le b$ . If m is a number between $f(a)$ and $f(b)$ , then there is a number c in the interval $a \le x \le b$ such that

$$f(c) = m.$$

The theorem is illustrated in the picture below:

$$\hbox{\epsfysize=1.75in \epsffile{continuity5.eps}}$$

Try it for yourself: Pick any height m between $f(a)$ and $f(b)$ . Move horizontally from your chosen height to the graph, then downward from the graph till you hit the x-axis. The place where you hit the x-axis is c. You'll always be able to do this if f is continuous. The intuitive idea is that, being continuous, f can't skip any values in going from $f(a)$ to $f(b)$ .

A proof of the Intermediate Value Theorem uses some deep properties of the real numbers, so I won't give it here. At least you can see from the picture that the result is geometrically reasonable.

The theorem illustrates an important point: You can know something exists without being able to find it.

If I take your car keys and throw them into a nearby corn field, you know that your keys are in the field --- but finding them is a different story!

The Intermediate Value Theorem says there is a number c such that $f(c) = m$ . It doesn't tell you how to find it, though you can usually approximate c as closely as you want.

And by the way --- there may be more than one number c which works. Even though the statement of the theorem says "there is" (singular), mathematicians use these words to mean "there is at least

Example. Suppose f is a continuous function, $f(4) = 11$ , and $f(7) = 2$ . Prove that for some number x between 4 and 7, $f(x) + x^2 = 42$ .

Since $x^2$ and $f(x)$ are continuous, $f(x) + x^2$ is continuous. Plug in 4 and 7:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & x & & $f(x) + x^2$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 4 & & $f(4) + 4^2 = 27$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 7 & & $f(7) + 7^2 = 51$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} }} $$

42 is between 27 and 51, so I can apply the Intermediate Value Theorem to $f(x) + x^2$ . It says that there is a number x between 4 and 7 such that $f(x) + x^2 = 42$ .

Example. Approximate a solution to the equation

$$x^5 + 7 x^3 + 7 x + 1 = 0.$$

Here's the graph:

$$\hbox{\epsfysize=2in \epsffile{continuity6.eps}}$$

It looks as though there's a root between -0.5 and 0.

A clever person might say at this point: "Why not just look up the general formula for solving a 5-th degree equation?" After all, there's the general quadratic formula for quadratics ... and there's a general cubic formula and a general quartic formula, though you'd probably have to look them up in a book of tables.

Unfortunately, you'll never find a general quintic formula in any book of tables. Nils Henrik Abel and and Paolo Ruffini showed almost 150 years ago that there's no general quintic, and Evariste Galois showed a little later that you won't have any luck with higher degree equations, either.

You can still approximate the root, and the Intermediate Value Theorem guarantees that there is one.

f (being a polynomial) is surely continuous. In this situation, the IVT says that f can't go from negative to positive without passing through 0 somewhere in between.

Notice that

$$f(-0.5) \approx -3.40625 \quad\hbox{and}\quad f(0) = 1.$$

Thus, I know there's a root between -0.5 and 0.

I'll approximate the root by bisection. At each step, I'll know the root is caught between two numbers. I'll plug the midpoint into f. The root is now on one side or the other, and I just keep going.

This is exactly what common sense would lead you to do.

Here's the computation:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & \cr & x & & $f(x)$ positive & & $f(x)$ negative & \cr height2 pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & \cr & -0.5 & & & & -3.40625 & \cr height2 pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & \cr & 0.0 & & 1 & & & \cr height2 pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & \cr & -0.25 & & & & -0.860352 & \cr height2 pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & \cr & -0.125 & & 0.111298 & & & \cr height2 pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & \cr & -0.1875 & & & & -0.358874 & \cr height2 pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2 pt & \omit & & \omit & & \omit & \cr & -0.15625 & & & & -0.120546 & \cr height2 pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

At this point, the root c is caught between -0.125 (the last x which made f positive) and -0.15625 (the last x which made f negative). These two numbers are 0.03125 apart. Hence, the midpoint $x = -0.140625$ is within $0.03125/2 = 0.015625$ of the actual root. The estimate $x = -0.140625$ is therefore good to within 1 or 2 one-hundredths.

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