Limits for Several Variables

The concepts of limits and continuity can be extended to the multivariable setting. In a one-variable limit, $g(x)$ approaches $L$ as $x$ approaches $a$ from only two possible directions: from the left or from the right:

In the case of two-variable functions however, the limit of $f(x, y)$ must approach $L$ no matter how $(x, y)$ approaches the point $(a, b)$ . Continuity in two-variable functions is straightforward from Definition 1.2:

A function $g(x)$ is said to be continuous at $x=a$ if,

\lim _{x \rightarrow a} g(x)=g(a) .

Further, the function is said to be continuous on an interval, $\mathcal{I}$ if it is continuous at every point on the interval.

we say that the function $f$ is continuous at $(x, y)=(a, b)$ if $f(x, y)$ approaches the function value $f(a, b)$ as $(x, y) \rightarrow(a, b)$ .

Limit Laws

Suppose $\lim _{(x, y) \rightarrow(a, b)} f(x, y)$ and $\lim _{(x, y) \rightarrow(a, b)} g(x, y)$ exist. Then:

Sum Law:

\lim _{(x, y) \rightarrow(a, b)}(f(x, y)+g(x, y))=\lim _{(x, y) \rightarrow(a, b)} f(x, y)+\lim _{(x, y) \rightarrow(a, b)} g(x, y)

Constant Multiple Law:

\lim _{(x, y) \rightarrow(a, b)} k f(x, y)=k \lim _{(x, y) \rightarrow(a, b)} f(x, y),

for any number $k$ .

Product Law:

\lim _{(x, y) \rightarrow(a, b)} f(x, y) g(x, y)=\left(\lim _{(x, y) \rightarrow(a, b)} f(x, y)\right)\left(\lim _{(x, y) \rightarrow(a, b)} g(x, y)\right)

Quotient Law:

\lim _{(x, y) \rightarrow(a, b)} \frac{f(x, y)}{g(x, y)}=\frac{\lim _{(x, y) \rightarrow(a, b)} f(x, y)}{\lim _{(x, y) \rightarrow(a, b)} g(x, y)}

provided that $\lim _{(x, y) \rightarrow(a, b)} g(x, y) \neq 0$ .

It follows from Theorem 1.1 that arbitrary sums, multiples and products of continuous functions are also continuous. Let us now look at an example of evaluating limits of a two-variable function.

Examples

Example 1.3 Show that $f(x, y)=\frac{2 x^{2} y-3 x y^{2}}{x^{2}+y^{2}+1}$ is continuous and evaluate $\lim _{(x, y) \rightarrow(1,2)} f(x, y)$ .

Solution The function $f(x, y)$ is continuous everywhere since it is a rational function whose denominator, $x^{2}+y^{2}+1$ is never zero. Now, as with functions of one variable, we can evaluate the limit of a continuous, two-variable function using substitution:

\lim _{(x, y) \rightarrow(1,2)} f(x, y)=\frac{4-12}{1+4+1}=-\frac{4}{3} .

Finally, for completeness, let us look at how we go about showing that a limit does not exist.

Example 1.4 Evaluate $\lim _{(x, y) \rightarrow(0,0)} f(x, y)=\frac{x^{3} y}{x^{6}+y^{2}}$ .

Solution The function $f(x, y)$ is not continuous at the point of interest since the denominator at $(0,0)$ is zero. This tells us that there is at least a chance that the limit does not exist. To show that a limit does in fact not exist, we need to prove that $f(x, y)$ approaches a different value as $(x, y) \rightarrow(0,0)$ along two different paths. We approach a point along the path by either fixing $x$ or $y$ or by relating $x$ or $y$ through some function (i.e. $y=x$ ). Let us first try the path $y=x$ [i.e. replace all the $y$ 's with the $x$ 's in $f(x, y)$ .

\lim _{(x, y) \rightarrow(0,0)} \frac{x^{3} y}{x^{6}+y^{2}}=\lim _{(x, x) \rightarrow(0,0)} \frac{x^{4}}{x^{2}\left(x^{4}+1\right)}=\lim _{(x, x) \rightarrow(0,0)} \frac{x^{2}}{x^{4}+1}=0

Now, try the path $y=x^{3}$ :

\lim _{(x, y) \rightarrow(0,0)} \frac{x^{3} y}{x^{6}+y^{2}}=\lim _{(x, x) \rightarrow(0,0)} \frac{x^{3} x^{3}}{x^{6}+x^{6}}=\lim _{(x, x) \rightarrow(0,0)} \frac{x^{6}}{2 x^{6}}=\lim _{(x, x) \rightarrow(0,0)} \frac{1}{2}=\frac{1}{2} \text {. }

We have two paths that give different values for the limit and hence the limit does not exist.

Limits Examples