Section 7.2 Limits and Continuity
¶To develop calculus for functions of one variable, we needed to make sense of the concept of a limit, which was used in the definition of a continuous function and the derivative of a function. Limits involving functions of two variables can be considerably more difficult to deal with; fortunately, most of the functions we encounter are fairly easy to understand.
The potential difficulty is largely due to the fact that there are many ways to “approach” a point in the \(x\)\(y\)plane. If we want to say that
we need to capture the idea that as \((x,y)\) gets close to \((a,b)\) then \(f(x,y)\) gets close to \(L\text{.}\) For functions of one variable, \(f(x)\text{,}\) there are only two ways that \(x\) can approach \(a\text{:}\) from the left or right. But there are an infinite number of ways to approach \((a,b)\text{:}\) along any one of an infinite number of straight lines, or even along a curved path in the \(x\)\(y\)plane. We might hope that it's really not so bad—suppose, for example, that along every possible line through \((a,b)\) the value of \(f(x,y)\) gets close to \(L\text{;}\) surely this means that “\(f(x,y)\) approaches \(L\) as \((x,y)\) approaches \((a,b)\)”. Sadly, no.
Example 7.11. Weird Limit.
Analyze \(f(x,y)=xy^2/(x^2+y^4)\text{.}\)
When \(x=0\) or \(y=0\text{,}\) \(f(x,y)\) is 0, so the limit of \(f(x,y)\) approaching the origin along either the \(x\) or \(y\) axis is 0. Moreover, along the line \(y=mx\text{,}\) \(f(x,y)=m^2x^3/(x^2+m^4x^4)\text{.}\) As \(x\) approaches 0 this expression approaches 0 as well. So along every line through the origin \(f(x,y)\) approaches 0. Now suppose we approach the origin along \(x=y^2\text{.}\) Then
so the limit is \(1/2\text{.}\) Looking at Figure 7.1, it is apparent that there is a ridge above \(x=y^2\text{.}\) Approaching the origin along a straight line, we go over the ridge and then drop down toward 0, but approaching along the ridge the height is a constant \(1/2\text{.}\)
Fortunately, we can define the concept of limit without needing to specify how a particular point is approached—indeed, in Definition 3.4, we didn't need the concept of “approach.” Roughly, that definition says that when \(x\) is close to \(a\) then \(f(x)\) is close to \(L\text{;}\) there is no mention of “how” we get close to \(a\text{.}\) We can adapt that definition to two variables quite easily:
Definition 7.12. Limit of a Multivariate Function.
Suppose \(f(x,y)\) is a twovariable function. We say that
if for every \(\epsilon>0\) there is a \(\delta > 0\) so that whenever \(0 \lt \sqrt{(xa)^2+(yb)^2} \lt \delta\text{,}\) \(f(x,y)L\lt \epsilon\text{.}\)
This says that we can make \(f(x,y)L\lt \epsilon\text{,}\) no matter how small \(\epsilon\) is, by making the distance from \((x,y)\) to \((a,b)\) “small enough”.
Example 7.13. Multivariate Limit.
Show that \(\ds \lim_{(x,y)\to(0,0)}{3x^2y\over x^2+y^2}=0\text{.}\)
Suppose \(\epsilon>0\text{.}\) Then
Note that \(x^2/(x^2+y^2)\le1\) and \(y=\sqrt{y^2}\le\sqrt{x^2+y^2}\lt \delta\text{.}\) So
We want to force this to be less than \(\epsilon\) by picking \(\delta\) “small enough.” If we choose \(\delta=\epsilon/3\) then
Recall that a function \(f(x)\) is continuous at \(x=a\) if \(\ds\lim_{x\to a}f(x)=f(a)\text{.}\) We can say exactly the same thing about a function of two variables: \(f(x,y)\) is continuous at \((a,b)\) if \(\ds\lim_{(x,y)\to (a,b)}f(x,y)=f(a,b)\text{.}\)
The function \(f(x,y)=3x^2y/(x^2+y^2)\) is not continuous at \((0,0)\text{,}\) because \(f(0,0)\) is not defined. However, we know that \(\ds \lim_{(x,y)\to(0,0)}f(x,y)=0\text{,}\) so we can make a continuous function, by extending the definition of \(f\) so that \(f(0,0)=0\text{.}\) This surface is shown in Figure 7.2.
Note that we cannot extend the definition of the function in Example 7.11 to create a continuous function, since the limit does not exist as we approach \((0,0)\text{.}\)
Fortunately, the functions we will be working with will usually be continuous almost everywhere. As with single variable functions, two classes of common functions are particularly useful and easy to describe. A polynomial in two variables is a sum of terms of the form \(ax^my^n\text{,}\) where \(a\) is a real number and \(m\) and \(n\) are nonnegative integers. A rational function is a quotient of polynomials.
Theorem 7.14. Continuity of Functions.
Polynomials are continuous everywhere. Rational functions are continuous everywhere they are defined.
Exercises for Section 7.2.
Exercise 7.2.1.
Determine whether each limit exists. If it does, find the limit and prove that it is the limit; if it does not, explain how you know.

\(\ds\lim_{(x,y)\to(0,0)}{x^2\over x^2+y^2}\)
AnswerNo limit; use \(x=0\) and \(y=0\) 
\(\ds\lim_{(x,y)\to(0,0)}{xy\over x^2+y^2}\)
AnswerNo limit; use \(x=0\) and \(x=y\) 
\(\ds\lim_{(x,y)\to(0,0)}{xy\over 2x^2+y^2}\)
AnswerNo limit; use \(x=0\) and \(x=y\) 
\(\ds\lim_{(x,y)\to(0,0)}{x^4y^4\over x^2+y^2}\)
AnswerLimit is 1 
\(\ds\lim_{(x,y)\to(0,0)}{\sin(x^2+y^2)\over x^2+y^2}\)
AnswerLimit is zero 
\(\ds\lim_{(x,y)\to(0,0)}{xy\over \sqrt{2x^2+y^2}}\)
AnswerLimit is zero 
\(\ds\lim_{(x,y)\to(0,0)} {e^{x^2y^2}1\over x^2+y^2}\)
AnswerLimit is \(1\) 
\(\ds\lim_{(x,y)\to(0,0)}{x^3+y^3\over x^2+y^2}\)
AnswerLimit is zero 
\(\ds\lim_{(x,y)\to(0,0)}{x^2 + \sin^2 y\over 2x^2+y^2}\)
AnswerNo limit; use \(x=0\) and \(y=0\) 
\(\ds\lim_{(x,y)\to(1,0)}{(x1)^2\ln x\over(x1)^2+y^2}\)
AnswerLimit is zero 
\(\ds\lim_{(x,y)\to(1,1)}{3x+4y}\)
AnswerLimit is \(1\) 
\(\ds\lim_{(x,y)\to(0,0)}{4x^2y\over x^2+y^2}\)
AnswerLimit is zero
Exercise 7.2.2.
Does the function \(\ds f(x,y)={xy\over 1+x+y}\) have any discontinuities? What about \(\ds f(x,y)={xy\over 1+x^2+y^2}\text{?}\) Explain.