Continuity Laws and Compositions

Why Continuity Laws Are Useful

Once continuity at a point has been defined, many new continuity arguments can be built from functions whose continuity is already known. Instead of returning to the definition for every formula, we can use laws for sums, products, quotients, and compositions.

These laws are efficient, but they are not unconditional. A quotient requires a nonzero denominator at the point. A composition requires the inner function to take values in the domain of the outer function, at least throughout some neighborhood of the point under consideration. Checking these requirements is part of the argument, not an optional technical detail.

Continuity at a Point

For a function $f:D\to\mathbb{R}^m$, where $D\subseteq\mathbb{R}^n$, continuity at $\mathbf{a}\in D$ means that $f(\mathbf{x})$ approaches $f(\mathbf{a})$ as $\mathbf{x}$ approaches $\mathbf{a}$ through points of $D$. Equivalently, $f$ is continuous at $\mathbf{a}$ when

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

The norms in the definition are the usual Euclidean norms in the corresponding spaces. In the scalar case, the output norm is absolute value. This tutorial uses continuity at a point together with the limit laws for multivariable functions.

Definition: A function $f:D\to\mathbb{R}^m$ is continuous at $\mathbf{a}\in D$ if, for every $\varepsilon>0$, there exists $\delta>0$ such that for every $\mathbf{x}\in D$, $$ \|\mathbf{x}-\mathbf{a}\|<\delta \quad\Longrightarrow\quad \|f(\mathbf{x})-f(\mathbf{a})\|<\varepsilon. $$

Continuity is always relative to the function’s domain. In particular, when a function is defined only on a subset of a Euclidean space, the points approaching $\mathbf{a}$ must belong to that subset.

Sum, Product, and Quotient Laws

Suppose $f$ and $g$ are scalar-valued functions defined on a common domain $D\subseteq\mathbb{R}^n$, and both are continuous at $\mathbf{a}\in D$. Their sum and product are continuous at $\mathbf{a}$. If $g(\mathbf{a})\ne 0$, their quotient is also continuous at $\mathbf{a}$, wherever that quotient is defined.

Theorem: If $f,g:D\to\mathbb{R}$ are continuous at $\mathbf{a}$, then $f+g$ and $fg$ are continuous at $\mathbf{a}$. If, in addition, $g(\mathbf{a})\ne 0$, then $f/g$ is continuous at $\mathbf{a}$.

The sum and product conclusions follow by applying the corresponding limit laws and the equality between the limit and the function value for a continuous function. For the quotient, the denominator condition matters: continuity of $g$ and $g(\mathbf{a})\ne0$ ensure that $g$ stays nonzero sufficiently close to $\mathbf{a}$. Indeed, continuity gives a neighborhood in which

|g(\mathbf{x})-g(\mathbf{a})|<\frac{|g(\mathbf{a})|}{2}.

The reverse triangle inequality then gives

|g(\mathbf{x})| \ge |g(\mathbf{a})|-|g(\mathbf{x})-g(\mathbf{a})| > \frac{|g(\mathbf{a})|}{2}>0.

Thus the quotient is defined nearby, and the quotient limit law applies. If $g(\mathbf{a})=0$, this law gives no conclusion: a particular quotient might still have a continuous extension, but that requires a separate argument.

Worked Example: A Polynomial Built from Continuity Laws

Consider $P(x,y)=x^2y+3x-2y$ at $\mathbf{a}=(1,2)$. Coordinate functions $x$ and $y$ are continuous, and sums and products of continuous functions are continuous. Therefore $x^2y+3x-2y$ is continuous at $(1,2)$.

Evaluate the function by substituting $x=1$ and $y=2$:

$$ P(1,2) =(1)^2(2)+3(1)-2(2) =2+3-4 =1. $$

The continuity claim says that $P(x,y)$ approaches $1$ as $(x,y)$ approaches $(1,2)$. Direct substitution checks the stated function value: the first term is $2$, the second is $3$, and the third is $-4$, so their sum is $1$.

Worked Example: A Quotient with a Nonzero Denominator

Let

Q(x,y)=\frac{x^2+2y}{x-y+2}, \qquad \mathbf{a}=(1,0).

The numerator and denominator are polynomials, so each is continuous at $(1,0)$. The denominator’s value there is

1-0+2=3\ne0.

The quotient law therefore proves that $Q$ is continuous at $(1,0)$. To see explicitly that the denominator stays nonzero nearby, write $x=1+h$ and $y=k$. If $\sqrt{h^2+k^2}<1$, then $|h|<1$ and $|k|<1$, so

$$ x-y+2=3+h-k>3-|h|-|k|>1. $$

Thus the denominator is positive in this neighborhood. Finally, substituting the point into the quotient gives

Q(1,0)=\frac{1^2+2(0)}{1-0+2} =\frac{1}{3}.

The numerator evaluates to $1$, the denominator to $3$, and their quotient is $1/3$. This agrees with the value to which $Q(x,y)$ tends by continuity.

Compositions: Domain Conditions and Continuity

A composition has the form $g\circ f$, meaning $(g\circ f)(\mathbf{x})=g(f(\mathbf{x}))$. For this expression to make sense, the values produced by $f$ must belong to the domain of $g$. For continuity at a point, it is not enough that $f(\mathbf{a})$ belongs to that domain: $f(\mathbf{x})$ must also belong to it for all $\mathbf{x}$ sufficiently close to $\mathbf{a}$ in the domain of $f$.

Theorem: Let $f:D\to\mathbb{R}^m$, let $g:E\to\mathbb{R}^p$, and let $\mathbf{a}\in D$. Suppose $f$ is continuous at $\mathbf{a}$, $f(\mathbf{a})\in E$, and there exists $\rho>0$ such that $f(\mathbf{x})\in E$ whenever $\mathbf{x}\in D$ and $\|\mathbf{x}-\mathbf{a}\|<\rho$. If $g$ is continuous at $f(\mathbf{a})$ as a function on $E$, then $g\circ f$ is continuous at $\mathbf{a}$.

Here is the full epsilon-delta justification. Fix any $\varepsilon>0$. Since $g$ is continuous at $f(\mathbf{a})$, there is an $\eta>0$ such that for every $\mathbf{u}\in E$,

\|\mathbf{u}-f(\mathbf{a})\|<\eta \quad\Longrightarrow\quad \|g(\mathbf{u})-g(f(\mathbf{a}))\|<\varepsilon.

Since $f$ is continuous at $\mathbf{a}$, there is a $\delta_1>0$ such that for $\mathbf{x}\in D$,

\|\mathbf{x}-\mathbf{a}\|<\delta_1 \quad\Longrightarrow\quad \|f(\mathbf{x})-f(\mathbf{a})\|<\eta.

The radius $\delta_1$ alone may not ensure $f(\mathbf{x})\in E$. The neighborhood condition in the theorem supplies a radius $\rho$ that does ensure it. Choose

\delta=\min(\delta_1,\rho).

If $\mathbf{x}\in D$ and $\|\mathbf{x}-\mathbf{a}\|<\delta$, then $\|\mathbf{x}-\mathbf{a}\|<\rho$, so $f(\mathbf{x})\in E$. Also $\|\mathbf{x}-\mathbf{a}\|<\delta_1$, so $\|f(\mathbf{x})-f(\mathbf{a})\|<\eta$. We may therefore substitute $\mathbf{u}=f(\mathbf{x})$ into the implication for $g$, because this $\mathbf{u}$ belongs to $E$. It follows that

\|g(f(\mathbf{x}))-g(f(\mathbf{a}))\|<\varepsilon.

This is precisely continuity of $g\circ f$ at $\mathbf{a}$. The minimum of the two radii handles both requirements at once: the input is close enough for continuity of $f$, and its image lies in the domain required for $g$.

Worked Example: A Composition with a Checked Inner Domain

Define $f:\mathbb{R}^2\to\mathbb{R}^2$ and $g:E\to\mathbb{R}$, where $E=\{(u,v)\in\mathbb{R}^2:v\ne0\}$, by

f(x,y)=(2+x,\,1+x^2+y^2), \qquad g(u,v)=\frac{u}{v}.

Both coordinate functions of $f$ are continuous, so $f$ is continuous at $(0,0)$. The function $g$ is continuous at every point of $E$ by the quotient law. For every $(x,y)$, the second coordinate of $f(x,y)$ satisfies

1+x^2+y^2\ge1>0.

Consequently $f(x,y)\in E$ everywhere, and the composition condition holds on every neighborhood of $(0,0)$. At the point,

f(0,0)=(2,1), \qquad g(f(0,0))=g(2,1)=\frac{2}{1}=2.

The composition is therefore continuous at $(0,0)$, with value $2$. Direct substitution into the composite formula gives

(g\circ f)(x,y)=\frac{2+x}{1+x^2+y^2}, \qquad (g\circ f)(0,0)=\frac{2+0}{1+0+0}=2.

The direct evaluation confirms the value obtained from the two stages of the composition, and the denominator check confirms that the formula is defined nearby.

A Common Composition Mistake

A frequent error is to check only that $f(\mathbf{a})$ is in the domain of $g$, then apply the composition law without checking nearby inputs. For a concrete example, let $f(x,y)=-x^2-y^2$, and let $g(t)=\sqrt{t}$ with domain $E=[0,\infty)$. At $(0,0)$, $f(0,0)=0\in E$. It is tempting to say that both functions are continuous there and hence their composition is continuous there.

That argument is invalid. If $(x,y)\ne(0,0)$, then $x^2+y^2>0$, and hence

$$ f(x,y)=-x^2-y^2<0. $$

Thus $f(x,y)\notin E$ at every other point, however close to $(0,0)$. The expression $g(f(x,y))$ is not real-valued there, so it does not define a composition on a neighborhood of $(0,0)$. The correction is to verify the inner-function range condition before using the theorem. In this example the composition is defined at the single point $(0,0)$, but that fact alone does not establish continuity there.

Key takeaway: For a quotient, check that the denominator is nonzero at the point. For a composition, check that the inner function’s values lie in the outer function’s domain throughout some neighborhood of the point.

Using the Laws Efficiently

A reliable continuity argument can be organized as a short sequence of checks. First identify the point and the functions being combined. Next verify that the component functions are continuous there. Then check the extra condition specific to the operation: a nonzero denominator for a quotient, or a suitable neighborhood mapping into the outer domain for a composition. Finally, evaluate the function at the point if its value is needed.

Identify the ingredients.
List the component functions and the point at which continuity is being considered.

Check their continuity.
Use known continuity of the component functions and the sum or product laws where appropriate.

Verify the operation’s condition.
For a quotient, check the denominator at the point. For a composition, check that the inner values belong to the outer domain nearby.

Apply the law and evaluate.
State the resulting continuity conclusion and substitute the point to confirm the function value.

These laws matter because complicated formulas are often assembled from simpler functions. Once continuity of the pieces and the relevant conditions are established, the continuity of the whole expression follows without a new epsilon-delta proof each time. The conditions also explain the limits of the method: if a denominator vanishes or a composition leaves its outer domain, the law cannot be applied as stated.

Check Your Understanding

Answer each question by stating both the continuity conclusion and any condition needed to justify it.

If $f$ and $g$ are scalar-valued and continuous at $\mathbf{a}$, what can be concluded about $f+g$ and $fg$?
Why does the quotient law require $g(\mathbf{a})\ne0$, and what does continuity then imply nearby?
In the composition theorem, why is it not always enough to know that $f(\mathbf{a})$ belongs to the domain of $g$?
In the composition proof, what purpose does choosing $\delta=\min(\delta_1,\rho)$ serve?
For $f(x,y)=x^2+y^2$ and $g(t)=\sqrt{t}$, does the inner-function domain condition hold near $(0,0)$? Explain.