Limit Laws for Functions of Several Variables

Why Limit Laws Matter

When a multivariable expression is built from simpler functions, its limit can often be found by combining the limits of those functions. The sum, scalar-multiple, product, and quotient laws make this precise. Their hypotheses matter: in particular, the quotient law requires the denominator’s limit to be nonzero.

These laws concern a function’s behavior as its input approaches a point, not necessarily its value at that point. The functions need not even be defined at the limiting point, provided they are defined sufficiently close to it for the limit to make sense.

Limit Definition and Common Domain

Let \(D\) be a subset of \(\mathbb{R}^n\), let \(\mathbf{a}\) be an accumulation point of \(D\), and let \(f:D\to\mathbb{R}\). Saying that \(f(\mathbf{x})\) approaches \(L\) as \(\mathbf{x}\) approaches \(\mathbf{a}\) means that values of \(f\) can be made arbitrarily close to \(L\) by taking \(\mathbf{x}\) sufficiently close to \(\mathbf{a}\), with \(\mathbf{x}\ne\mathbf{a}\).

Definition: We write \(\displaystyle\lim_{\mathbf{x}\to\mathbf{a}} f(\mathbf{x})=L\) if, for every \(\varepsilon>0\), there is a \(\delta>0\) such that whenever \(\mathbf{x}\in D\) and \(0<\|\mathbf{x}-\mathbf{a}\|<\delta\), we have \(|f(\mathbf{x})-L|<\varepsilon\). Here \(\|\cdot\|\) is the Euclidean norm.

For a limit law involving two functions \(f\) and \(g\), both functions must be defined on a common set containing all points of \(D\) sufficiently close to \(\mathbf{a}\), except possibly \(\mathbf{a}\) itself. Their limits are taken as \(\mathbf{x}\to\mathbf{a}\) within that common set. This ensures that the sums and products under discussion are defined near the point.

Suppose, under these conditions, that \(\lim_{\mathbf{x}\to\mathbf{a}}f(\mathbf{x})=A\) and \(\lim_{\mathbf{x}\to\mathbf{a}}g(\mathbf{x})=B\), where \(A\) and \(B\) are real numbers.

Theorem: Limit laws. Under the common-domain conditions above:

\(\displaystyle\lim_{\mathbf{x}\to\mathbf{a}}(f(\mathbf{x})+g(\mathbf{x}))=A+B\).
For any constant \(c\in\mathbb{R}\), \(\displaystyle\lim_{\mathbf{x}\to\mathbf{a}}cf(\mathbf{x})=cA\).
\(\displaystyle\lim_{\mathbf{x}\to\mathbf{a}}f(\mathbf{x})g(\mathbf{x})=AB\).
If \(B\ne 0\), and \(g(\mathbf{x})\ne 0\) sufficiently close to \(\mathbf{a}\) on the common domain, then \(\displaystyle\lim_{\mathbf{x}\to\mathbf{a}}\frac{f(\mathbf{x})}{g(\mathbf{x})}=\frac{A}{B}\).

The nonzero-limit condition in the quotient law ensures the denominator stays away from zero sufficiently near the limiting point. The other laws do not require their limiting values to be nonzero.

Why the Laws Hold

The sum and scalar-multiple laws follow directly from the definition. For the sum, given \(\varepsilon>0\), choose neighborhoods in which \(|f(\mathbf{x})-A|<\varepsilon/2\) and \(|g(\mathbf{x})-B|<\varepsilon/2\). In the smaller of these neighborhoods, the triangle inequality gives

|(f(\mathbf{x})+g(\mathbf{x}))-(A+B)| \le |f(\mathbf{x})-A|+|g(\mathbf{x})-B| <\varepsilon.

For a scalar \(c\ne 0\), choose a neighborhood in which \(|f(\mathbf{x})-A|<\varepsilon/|c|\). Then \(|cf(\mathbf{x})-cA|<\varepsilon\). If \(c=0\), both \(cf(\mathbf{x})\) and \(cA\) are zero, so the conclusion holds immediately.

The product requires one extra observation: a function with a finite limit is bounded sufficiently close to the limiting point. Here is a complete proof that uses this fact.

Bound \(f\) near the point.
Since \(f(\mathbf{x})\to A\), choose a neighborhood where \(|f(\mathbf{x})-A|<1\). The triangle inequality then gives \(|f(\mathbf{x})|<|A|+1\).

Split the product difference.
For every point in the common domain, \(f(\mathbf{x})g(\mathbf{x})-AB=f(\mathbf{x})(g(\mathbf{x})-B)+B(f(\mathbf{x})-A)\).

Choose bounds for a given tolerance.
For \(\varepsilon>0\), take a neighborhood where \(|g(\mathbf{x})-B|<\varepsilon/[2(|A|+1)]\) and \(|f(\mathbf{x})-A|<\varepsilon/[2(|B|+1)]\), as well as the bound from Step 1.

Conclude the estimate.
The triangle inequality bounds the product difference by a quantity less than \(\varepsilon/2+\varepsilon/2=\varepsilon\). Thus the product tends to \(AB\).

To justify the quotient law, first consider the reciprocal. Since \(g(\mathbf{x})\to B\ne 0\), choose a neighborhood where \(|g(\mathbf{x})-B|<|B|/2\). The reverse triangle inequality gives \(|g(\mathbf{x})|>|B|/2\), so \(g(\mathbf{x})\ne0\) there. Moreover,

\left|\frac{1}{g(\mathbf{x})}-\frac{1}{B}\right| =\frac{|g(\mathbf{x})-B|}{|g(\mathbf{x})||B|} <\frac{2|g(\mathbf{x})-B|}{|B|^2}.

The right-hand side can be made arbitrarily small because \(g(\mathbf{x})\to B\). Therefore \(1/g(\mathbf{x})\to1/B\). Applying the product law to \(f(\mathbf{x})\) and \(1/g(\mathbf{x})\) yields \(f(\mathbf{x})/g(\mathbf{x})\to A/B\), as claimed.

Worked Examples

Worked Example: Combining Limits by Addition and Scalar Multiplication

Let \(f(x,y)=2+x-2y\) and \(g(x,y)=3x+y\). Find the limit of \(3f(x,y)-2g(x,y)\) as \((x,y)\to(1,1)\).

First, the Euclidean distance bound gives \(|x-1|\le\|(x,y)-(1,1)\|\) and \(|y-1|\le\|(x,y)-(1,1)\|\). Therefore,

|f(x,y)-1| =|(x-1)-2(y-1)| \le 3\|(x,y)-(1,1)\|,

so \(f(x,y)\to1\). Similarly,

|g(x,y)-4| =|3(x-1)+(y-1)| \le 4\|(x,y)-(1,1)\|,

so \(g(x,y)\to4\). The scalar-multiple and sum laws now give

\lim_{(x,y)\to(1,1)}(3f(x,y)-2g(x,y)) =3(1)-2(4)=-5.

Check: expanding the expression gives \(3f-2g=6-3x-8y\). At \((1,1)\), this expression has value \(6-3-8=-5\), matching the limit obtained from the laws.

Worked Example: Applying the Product Law

Let \(f(x,y)=1+x-y\) and \(g(x,y)=2+2x+y\). Find \(\lim_{(x,y)\to(1,1)}f(x,y)g(x,y)\).

Writing \(x-1\) and \(y-1\) as the coordinate differences gives

|f(x,y)-1|=|(x-1)-(y-1)|\le2\|(x,y)-(1,1)\|,

and

|g(x,y)-5|=|2(x-1)+(y-1)|\le3\|(x,y)-(1,1)\|.

Thus \(f(x,y)\to1\) and \(g(x,y)\to5\). Both limits are finite, so the product law applies and gives

\lim_{(x,y)\to(1,1)}f(x,y)g(x,y)=1\cdot5=5.

Check: at \((1,1)\), the two factors have values \(1+1-1=1\) and \(2+2+1=5\). Their product is \(1\cdot5=5\), agreeing with the calculated limit.

Worked Example: Applying the Quotient Law

Let \(f(x,y)=1+x+y\) and \(g(x,y)=4+x-2y\). Find \(\lim_{(x,y)\to(0,0)}f(x,y)/g(x,y)\).

Using \(|x|\le\|(x,y)\|\) and \(|y|\le\|(x,y)\|\), we obtain

|f(x,y)-1|=|x+y|\le2\|(x,y)\|, \qquad |g(x,y)-4|=|x-2y|\le3\|(x,y)\|.

Hence \(f(x,y)\to1\) and \(g(x,y)\to4\). The denominator limit is \(4\ne0\). Also, whenever \(\|(x,y)\|<1\),

|g(x,y)-4|<3 \qquad\Longrightarrow\qquad |g(x,y)|>1,

so the denominator is nonzero throughout that neighborhood. The quotient law applies:

\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{g(x,y)} =\frac{1}{4}.

Check: the numerator and denominator limits used in the quotient are \(1\) and \(4\), respectively, and their ratio is \(1/4\). The denominator is also bounded away from zero near the point, as the law requires.

A Common Quotient-Law Mistake

A frequent error is to apply the quotient law when the denominator’s limit is zero. For example, let \(f(x,y)=x^2+y^2\) and \(g(x,y)=x^2+y^2\), and consider their quotient away from \((0,0)\). Both numerator and denominator tend to zero. The incorrect calculation is:

\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{g(x,y)} \overset{\text{incorrect}}{=} \frac{0}{0}.

This is not a valid application of the quotient law: its denominator-limit hypothesis fails, and \(0/0\) is not a real number. In this particular example, the quotient is defined at every point other than the origin and satisfies

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

Consequently, its limit is \(1\), not \(0/0\). The lesson is precise: a zero denominator limit means the quotient law cannot be used; it does not, by itself, decide whether the quotient limit exists.

Using the Laws Carefully

Before applying a limit law, identify the component limits and check that the functions share a suitable domain near the point. For a quotient, check both that the denominator limit is nonzero and that the denominator function is nonzero sufficiently close to the point on that domain. When the denominator limit is nonzero, the latter condition follows from the proof: the denominator is then bounded away from zero nearby.

The laws give limits for expressions assembled from functions whose limits are already known. They do not say that every expression with familiar-looking algebra has a limit, nor do they permit arithmetic with undefined quantities such as \(0/0\). Stating the hypotheses alongside the calculation is part of using the laws correctly.

Check Your Understanding

Use the hypotheses of each law as well as its calculation when answering.

If \(f(\mathbf{x})\to -2\) and \(g(\mathbf{x})\to 5\), what is the limit of \(4f(\mathbf{x})+g(\mathbf{x})\)?
What extra estimate is needed in the product-law proof before bounding \(f(\mathbf{x})(g(\mathbf{x})-B)\)?
If \(f(\mathbf{x})\to3\) and \(g(\mathbf{x})\to0\), may the quotient law be used to find the limit of \(f(\mathbf{x})/g(\mathbf{x})\)? Explain.
Why does \(g(\mathbf{x})\to B\ne0\) ensure that \(g(\mathbf{x})\) is nonzero sufficiently close to the limiting point?
If both the numerator and denominator tend to zero, what can you conclude from the quotient law alone about the quotient limit?