## Section7.2Limits 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.

###### Example7.11. 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 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:

###### Definition7.12. 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”.

###### Example7.13. 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 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 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. $\ds\lim_{(x,y)\to(0,0)}{x^2\over x^2+y^2}$

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

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

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

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

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

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

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

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

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

11. $\ds\lim_{(x,y)\to(1,-1)}{3x+4y}$
Limit is $-1$
12. $\ds\lim_{(x,y)\to(0,0)}{4x^2y\over x^2+y^2}$
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.