Mathematics · Ch 5 — Continuity and Differentiability

Algebra of Continuous Functions

5.2.1

Algebra of Continuous Functions

5.2.1 Algebra of Continuous Functions

Since continuity at a point is defined entirely by the limit at that point, continuous functions inherit the algebra of limits. This gives powerful tools to build new continuous functions from known ones.

Theorem 1: Algebraic Operations Preserve Continuity

Suppose $f$ and $g$ are two real functions that are both continuous at a real number $c$ . Then the following functions are also continuous at $x = c$ :

$f + g$ (the sum)
$f - g$ (the difference)
$f \cdot g$ (the product)
$\frac{f}{g}$ (the quotient), provided $g(c) \neq 0$

Important

These four results mirror the algebra of limits exactly. The reason: continuity at $c$ means $\lim_{x \to c} f(x) = f(c)$ and $\lim_{x \to c} g(x) = g(c)$ , so the limit laws apply directly.

Proof of (1): Continuity of $f + g$

The function $f + g$ is defined at $x = c$ by $(f + g)(c) = f(c) + g(c)$ , a real number since both $f(c)$ and $g(c)$ exist. Now compute the limit:

\begin{aligned} \lim_{x \to c} (f + g)(x) &= \lim_{x \to c} [f(x) + g(x)] \quad \text{(definition of } f + g) \\ &= \lim_{x \to c} f(x) + \lim_{x \to c} g(x) \quad \text{(limit of a sum)} \\ &= f(c) + g(c) \quad \text{(continuity of } f \text{ and } g) \\ &= (f + g)(c) \end{aligned}

Since $\lim_{x \to c} (f + g)(x) = (f + g)(c)$ , the function $f + g$ is continuous at $x = c$ .

›Proof

Proof of (2): Continuity of $f - g$

$\begin{aligned} \lim_{x \to c} (f - g)(x) &= \lim_{x \to c} [f(x) - g(x)] \\ &= \lim_{x \to c} f(x) - \lim_{x \to c} g(x) \quad \text{(limit of a difference)} \\ &= f(c) - g(c) \quad \text{(continuity of } f \text{ and } g) \\ &= (f - g)(c) \end{aligned}$

Hence $f - g$ is continuous at $x = c$ .

›Proof

Proof of (3): Continuity of $f \cdot g$

$\begin{aligned} \lim_{x \to c} (f \cdot g)(x) &= \lim_{x \to c} [f(x) \cdot g(x)] \\ &= \left( \lim_{x \to c} f(x) \right) \cdot \left( \lim_{x \to c} g(x) \right) \quad \text{(limit of a product)} \\ &= f(c) \cdot g(c) \quad \text{(continuity of } f \text{ and } g) \\ &= (f \cdot g)(c) \end{aligned}$

Therefore $f \cdot g$ is continuous at $x = c$ .

›Proof

Proof of (4): Continuity of $\frac{f}{g}$ (provided $g(c) \neq 0$ )

Since $g(c) \neq 0$ , the quotient is defined at $x = c$ by $\left(\frac{f}{g}\right)(c) = \frac{f(c)}{g(c)}$ .

$\begin{aligned} \lim_{x \to c} \left(\frac{f}{g}\right)(x) &= \lim_{x \to c} \frac{f(x)}{g(x)} \\ &= \frac{\displaystyle \lim_{x \to c} f(x)}{\displaystyle \lim_{x \to c} g(x)} \quad \text{(limit of a quotient, denominator non-zero)} \\ &= \frac{f(c)}{g(c)} \quad \text{(continuity of } f \text{ and } g) \\ &= \left(\frac{f}{g}\right)(c) \end{aligned}$

Hence $\frac{f}{g}$ is continuous at $x = c$ .

Watch out

The condition $g(c) \neq 0$ is essential. If $g(c) = 0$ , the quotient may still be continuous at $c$ in special cases (e.g., after cancellation), but Theorem 1 does not guarantee it. Always check the denominator at the point of interest.

Important Special Cases

Special Case of (3): Multiplying by a Constant

If $f$ is a constant function $f(x) = \lambda$ , then property (3) gives that $(\lambda \cdot g)(x) = \lambda \cdot g(x)$ is continuous wherever $g$ is continuous. In particular, taking $\lambda = -1$ , the continuity of $g$ implies the continuity of $-g$ .

Special Case of (4): Constant Numerator

If $f(x) = \lambda$ , then property (4) gives that $\left(\frac{\lambda}{g}\right)(x) = \frac{\lambda}{g(x)}$ is continuous wherever $g(x) \neq 0$ . In particular, taking $\lambda = 1$ , the continuity of $g$ implies the continuity of $\frac{1}{g}$ at all points where $g(x) \neq 0$ .

Tip

These special cases are practical: instead of reproving continuity for every scaled or reciprocal function, note that they are built from continuous functions using the algebra rules.

Composition of Continuous Functions

Theorem 2: Continuity of Composite Functions

Suppose $f$ and $g$ are real-valued functions such that $(f \circ g)$ is defined at $c$ . If $g$ is continuous at $c$ and $f$ is continuous at $g(c)$ , then $(f \circ g)$ is continuous at $c$ . …

Theorem 1

Theorem 2: Continuity of Composite Functions

Suppose $f$ and $g$ are real-valued functions such that the composite function $(f \circ g)$ is defined at $c$ . If $g$ is continuous at $c$ and $f$ is continuous at $g(c)$ , then the composite function $f \circ g$ is continuous at $c$ .

In symbols:

If $\lim_{x \to c} g(x) = g(c)$ and $\lim_{t \to g(c)} f(t) = f(g(c))$ , then $\lim_{x \to c} f(g(x)) = f(g(c))$ .

Important

The theorem requires two continuity checks: the inner function at $c$ , and the outer function at the value $g(c)$ . Both must hold for the composite to be continuous at $c$ .

Complete Proof

›Proof

Step 1: Set up the limit we need to evaluate.

We want to show $\lim_{x \to c} (f \circ g)(x) = (f \circ g)(c)$ .

By definition, $(f \circ g)(x) = f(g(x))$ and $(f \circ g)(c) = f(g(c))$ .

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

Step 2: Use the continuity of $g$ at $c$ .

Since $g$ is continuous at $c$ , we have $\lim_{x \to c} g(x) = g(c)$ .

This means: as $x$ approaches $c$ , the values $g(x)$ approach $g(c)$ .

Step 3: Use the continuity of $f$ at $g(c)$ .

Since $f$ is continuous at $g(c)$ , we have $\lim_{t \to g(c)} f(t) = f(g(c))$ .

This means: whenever $t$ approaches $g(c)$ , the values $f(t)$ approach $f(g(c))$ .

Step 4: Combine the two limits.

Let $t = g(x)$ . As $x \to c$ , we know $t \to g(c)$ (from Step 2).

Then $\lim_{x \to c} f(g(x)) = \lim_{t \to g(c)} f(t)$ (by substituting $t = g(x)$ ).

And $\lim_{t \to g(c)} f(t) = f(g(c))$ (from Step 3).

Step 5: Conclude.

Therefore $\lim_{x \to c} f(g(x)) = f(g(c))$ , which is exactly $\lim_{x \to c} (f \circ g)(x) = (f \circ g)(c)$ .

Hence $f \circ g$ is continuous at $c$ . $\square$

Note

The substitution $t = g(x)$ in Step 4 is valid because $g$ is continuous at $c$ — this guarantees that as $x$ gets arbitrarily close to $c$ , $g(x)$ gets arbitrarily close to $g(c)$ , so the limit of $f$ as $t \to g(c)$ captures exactly what happens to $f(g(x))$ as $x \to c$ .

When Is This Theorem Used? …

Theorem 2

Theorem 2: Continuity of Composite Functions

In symbols:

If $\lim_{x \to c} g(x) = g(c)$ and $\lim_{t \to g(c)} f(t) = f(g(c))$ , then $\lim_{x \to c} f(g(x)) = f(g(c))$ .

Important

The theorem requires two continuity checks: the inner function at $c$ , and the outer function at the value $g(c)$ . Both must hold for the composite to be continuous at $c$ .

Complete Proof

›Proof

Step 1: Set up the limit we need to evaluate.

We want to show $\lim_{x \to c} (f \circ g)(x) = (f \circ g)(c)$ .

By definition, $(f \circ g)(x) = f(g(x))$ and $(f \circ g)(c) = f(g(c))$ .

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

Step 2: Use the continuity of $g$ at $c$ .

Since $g$ is continuous at $c$ , we have $\lim_{x \to c} g(x) = g(c)$ .

This means: as $x$ approaches $c$ , the values $g(x)$ approach $g(c)$ .

Step 3: Use the continuity of $f$ at $g(c)$ .

Since $f$ is continuous at $g(c)$ , we have $\lim_{t \to g(c)} f(t) = f(g(c))$ .

This means: whenever $t$ approaches $g(c)$ , the values $f(t)$ approach $f(g(c))$ .

Step 4: Combine the two limits.

Let $t = g(x)$ . As $x \to c$ , we know $t \to g(c)$ (from Step 2).

Then $\lim_{x \to c} f(g(x)) = \lim_{t \to g(c)} f(t)$ (by substituting $t = g(x)$ ).

And $\lim_{t \to g(c)} f(t) = f(g(c))$ (from Step 3).

Step 5: Conclude.

Therefore $\lim_{x \to c} f(g(x)) = f(g(c))$ , which is exactly $\lim_{x \to c} (f \circ g)(x) = (f \circ g)(c)$ .

Hence $f \circ g$ is continuous at $c$ . $\square$

Note

When Is This Theorem Used? …

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Algebra of Continuous Functions