԰AV

Skip to main content

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

\begin{equation*} \ds\lim_{(x,y)\to(a,b)}f(x,y)=L\text{,} \end{equation*}

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.

Figure 7.1. \(\ds f(x,y)={xy^2\over x^2+y^4}\)
Example 7.11. Weird Limit.

Analyze \(f(x,y)=xy^2/(x^2+y^4)\text{.}\)

Solution

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

\begin{equation*} f(x,y)={y^2y^2\over y^4+y^4}={y^4\over2y^4}={1\over2}\text{,} \end{equation*}

so the limit is \(1/2\text{.}\) Looking at 󾱲ܰ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 ٱھԾپDz3.4, 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 two-variable function. We say that

\begin{equation*} \lim_{(x,y)\to (a,b)}f(x,y)=L \end{equation*}

if for every \(\epsilon>0\) there is a \(\delta > 0\) so that whenever \(0 \lt \sqrt{(x-a)^2+(y-b)^2} \lt \delta\text{,}\) \(|f(x,y)-L|\lt \epsilon\text{.}\)

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 \(\ds \lim_{(x,y)\to(0,0)}{3x^2y\over x^2+y^2}=0\text{.}\)

Solution

Suppose \(\epsilon>0\text{.}\) Then

\begin{equation*} \left|{3x^2y\over x^2+y^2}\right|={x^2\over x^2+y^2}3|y|\text{.} \end{equation*}

Note that \(x^2/(x^2+y^2)\le1\) and \(|y|=\sqrt{y^2}\le\sqrt{x^2+y^2}\lt \delta\text{.}\) So

\begin{equation*} {x^2\over x^2+y^2}3|y|\lt 1\cdot 3\cdot \delta\text{.} \end{equation*}

We want to force this to be less than \(\epsilon\) by picking \(\delta\) “small enough.” If we choose \(\delta=\epsilon/3\) then

\begin{equation*} \left|{3x^2y\over x^2+y^2}\right|\lt 1\cdot 3\cdot{\epsilon\over3}= \epsilon\text{.} \end{equation*}

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 󾱲ܰ7.2.

Figure 7.2. \(\ds f(x,y)={3x^2y\over x^2+y^2}\)

Note that we cannot extend the definition of the function in 油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 non-negative integers. A rational function is a quotient of polynomials.

Exercises for Section 7.2.

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.

  1. \(\ds\lim_{(x,y)\to(0,0)}{x^2\over x^2+y^2}\)

    Answer
    No limit; use \(x=0\) and \(y=0\)
  2. \(\ds\lim_{(x,y)\to(0,0)}{xy\over x^2+y^2}\)

    Answer
    No limit; use \(x=0\) and \(x=y\)
  3. \(\ds\lim_{(x,y)\to(0,0)}{xy\over 2x^2+y^2}\)

    Answer
    No limit; use \(x=0\) and \(x=y\)
  4. \(\ds\lim_{(x,y)\to(0,0)}{x^4-y^4\over x^2+y^2}\)

    Answer
    Limit is 1
  5. \(\ds\lim_{(x,y)\to(0,0)}{\sin(x^2+y^2)\over x^2+y^2}\)

    Answer
    Limit is zero
  6. \(\ds\lim_{(x,y)\to(0,0)}{xy\over \sqrt{2x^2+y^2}}\)

    Answer
    Limit is zero
  7. \(\ds\lim_{(x,y)\to(0,0)} {e^{-x^2-y^2}-1\over x^2+y^2}\)

    Answer
    Limit is \(-1\)
  8. \(\ds\lim_{(x,y)\to(0,0)}{x^3+y^3\over x^2+y^2}\)

    Answer
    Limit is zero
  9. \(\ds\lim_{(x,y)\to(0,0)}{x^2 + \sin^2 y\over 2x^2+y^2}\)

    Answer
    No limit; use \(x=0\) and \(y=0\)
  10. \(\ds\lim_{(x,y)\to(1,0)}{(x-1)^2\ln x\over(x-1)^2+y^2}\)

    Answer
    Limit is zero
  11. \(\ds\lim_{(x,y)\to(1,-1)}{3x+4y}\)

    Answer
    Limit is \(-1\)
  12. \(\ds\lim_{(x,y)\to(0,0)}{4x^2y\over x^2+y^2}\)

    Answer
    Limit is zero

Does the function \(\ds f(x,y)={x-y\over 1+x+y}\) have any discontinuities? What about \(\ds f(x,y)={x-y\over 1+x^2+y^2}\text{?}\) Explain.