# Continuous Function and Multivariable Limit • By David A. Smith, Founder & CEO, Direct Knowledge
• David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

## Intuitive Meaning of Multivariable Limit

Recall when considering $$\lim_{x\to c} f(x)=L$$ we need to examine the approach of $x$ to $c$ from two directions –namely the left-hand limit and the right-hand limit. However for functions of two variables, we write $(x,y)\to (a,b)$ to mean that the point $(x,y)$ is allowed to approach $(a,b)$ along any path in the domain of $f$ that passes through $(a,b).$

Open and closed disks are analogous to open and closed intervals on a coordinate line. An open disk (or open ball) centered at the point $(a,b)$ is the set of all points $(x,y)$ such that \begin{equation} \sqrt{(x-a)^2+(y-b)^2} < r \end{equation} for $r>0.$ Open disk centered at point $(a,b)$ with radius $r.$

If the boundary of the disk is included, the disk is called a closed disk. Closed disk centered at point $(a,b)$ with radius $r.$

A point $(a,b)$ is called an interior point of a set $S$ in $\mathbb{R}^2$ if some open disk centered at $(a,b)$ is contained entirely within $S.$ If $S$ is the empty set, or if every point of $S$ is an interior point, then $S$ is called an open set. The point $(a,b)$ is called a boundary point of $S$ if every open disk centered at $(a,b)$ contains both points that belong to $S$ and points that do not. The collection of all boundary points of $S$ is called the boundary of $S$, and $S$ is a closed disk if it contains its boundary.

Definition. Let $f$ be a function that is defined for all points $(x,y)$ in an open disk around $(a,b)$ with the possible exception of $(a,b)$ itself. Then the limit of $f(x,y)$ as $(x,y)$ approaches $(a,b)$ is $L$, written by \begin{equation} \lim_{(x,y)\to(a,b)} f(x,y)=L \end{equation} if $f(x,y)$ can be made as close to $L$ as we please by restricting $(x,y)$ to be sufficiently close to $(a,b).$ The intuitive meaning of the limit $L$ of a two variable function $f(x,y)$ at a point $(a,b)$.

## Limit Properties

If $a$, $b$, and $c$ are real numbers then \begin{equation} \label{linmultlimit} \lim_{(x,y)\to(a,b)}c=c,\qquad \lim_{(x,y)\to(a,b)}x=a, \qquad \lim_{(x,y)\to(a,b)}y =b. \end{equation} The three limits in \eqref{linmultlimit} can be used to show that \begin{equation} \label{polratlimit} \lim_{(x,y) \to(a,b)}p(x,y)=p(a,b) \qquad \text{and} \qquad \lim_{(x,y)\to(a,b)} r(x,y)=r(a,b) \end{equation} where $p$ is a polynomial function of $x$ and $y$ and $r$ is a rational function of $x$ and $y$, provided $(a,b)$ is in the domain of $r.$

Theorem. (Limit Laws) Suppose $$\lim_{(x,y)\to \left(x_0,y_0\right)}f(x,y)=L \quad \text{and} \quad \lim_{(x,y)\to \left(x_0,y_0\right)}g(x,y)=M$$ and $a$ is a real number. Then

$(1) \quad$ $\displaystyle \lim_{(x,y)\to \left(x_0,y_0\right)}(a f)(x,y)=a L$

$(2) \quad$ $\displaystyle \lim_{(x,y)\to \left(x_0,y_0\right)}(f+g)(x,y)=L+M$

$(3) \quad$ $\displaystyle \lim_{(x,y)\to \left(x_0,y_0\right)}(f g)(x,y)=L M$

$(4) \quad$ $\displaystyle \lim_{(x,y)\to \left(x_0,y_0\right)}(f/g)(x,y)=\frac{L}{M}$

whenever $M\neq 0.$

Example. Evaluate the following limit $\displaystyle \lim_{(x,y)\to (1,1)}\dfrac{x^4+y^4}{x^2+y^2}.$ A sketch of the function $f(x,y)=\dfrac{x^4+y^4}{x^2+y^2}.$

Solution. Since $r(x,y)=(x^4+y^4)/(x^2+y^2)$ is a rational function and $(1,1)$ is in the domain of $r$, we find \begin{equation} \lim_{(x,y)\to (1,1)}\dfrac{x^4+y^4}{x^2+y^2} =r(2,2)=1. \end{equation}

Example. Evaluate the following limit $\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{x^4-y^4}{x^2+y^2}.$ A sketch of the function $f(x,y)=\dfrac{x^4-y^4}{x^2+y^2}.$

Solution. By canceling \begin{equation} \label{cancellim} \lim_{(x,y)\to (0,0)}\frac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{x^2+y^2}=\lim_{(x,y)\to (0,0)}\left(x^2-y^2\right)=0. \end{equation} Notice the functions $f$ and $g$ defined by \begin{equation} f(x,y)=x^2-y^2 \qquad \text{and}\qquad g(x,y)=\dfrac{x^4-y^4}{x^2+y^2} \end{equation} are not the same functions, because they have different domains. However $f$ and $g$ have the same limit because whenever $(x,y)\neq(0,0)$ we do indeed have $(x^4-y^4)/(x^2+y^2)=x^2-y^2.$ Therefore, the canceling step in \eqref{cancellim} is valid.

Example. Evaluate the following limit $\displaystyle \lim_{(x,y)\to (1,2)}\dfrac{\left(x^2-1\right)\left(y^2-4\right)}{(x-1)(y-2)}.$ A sketch of the function $f(x,y)=\dfrac{\left(x^2-1\right)\left(y^2-4\right)}{(x-1)(y-2)}.$

Solution. By factoring and then canceling we have \begin{align} \lim_{(x,y)\to (1,2)}\frac{\left(x^2-1\right)\left(y^2-4\right)}{(x-1)(y-2)} & =\lim_{(x,y)\to (1,2)}\frac{(x-1)(x+1)(y-2)(y+2)}{(x-1)(y-2)} \\ & =\lim_{(x,y)\to (1,2)}(x+1)(y+2) =(1+1)(2+2) =8. \end{align}

Example. Evaluate the following limit $\displaystyle \lim_{(x,y)\to (a,a)}\dfrac{x^4-y^4}{x^2-y^2}.$ A sketch of the function $f(x,y)=\dfrac{x^4-y^4}{x^2-y^2}.$

Solution. By factoring and then canceling we have \begin{align} \lim_{(x,y)\to (a,a)}\frac{x^4-y^4}{x^2-y^2} & =\lim_{(x,y)\to (a,a)}\frac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{x^2-y^2} \\ & =\lim_{(x,y)\to (a,a)}\left(x^2+y^2\right) = 2a^2. \end{align}

## Limits that Do Not Exist

It is sometimes possible to show that the limit of a function does not exist by showing that the limit has different values depending on which path in the domain is used.

Theorem. (Limits Along Paths) If $f(x,y)$ approaches two different numbers as $(x,y)$ approaches $(a,b)$ along two different paths, then the limit $$\lim_{(x,y)\to(a,b)}f(x,y)$$ does not exist.

Example. Evaluate the following limit $\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{x y}{x^2+y^2}.$ A sketch of the function $f(x,y)=\dfrac{x y}{x^2+y^2}.$

Solution. Along the $x$-axis we have \begin{equation} \lim_{\begin{array}{c} (x,y)\to (0,0) \\ y=0 \end{array}} \dfrac{x y}{x^2+y^2} =\lim_{(x,0)\to (0,0)}\frac{0}{x^2+0^2}=\lim_{x\to 0}0=0.\end{equation} We have along the $y$-axis, \begin{equation} \lim_{ \begin{array}{c} (x,y)\to (0,0) \\ x=0 \end{array} }\dfrac{x y}{x^2+y^2} =\lim_{(0,y)\to (0,0)}\frac{0}{0^2+y^2}=\lim_{y\to 0}0=0.\end{equation} Along the line $y=x$ we have \begin{equation} \lim_{ \begin{array}{c} (x,y)\to (0,0) \\ y=x \end{array} }\dfrac{x y}{x^2+y^2} =\lim_{(x,x)\to (0,0)}\frac{ x^2 }{x^2+x^2}=\lim_{x\to 0}\frac{1}{2}=\frac{1}{2}. \end{equation} Therefore the given limit does not exist.

Example. Evaluate the following limit $\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{x y^2}{x^2+y^4}.$ A sketch of the function $f(x,y)=\dfrac{x y^2}{x^2+y^4}.$

Solution. Along the curves $y=m x$ we have \begin{equation} \lim_{ \begin{array}{c} (x,y)\to (0,0) \\ y=mx \end{array} }\dfrac{x y^2}{x^2+y^4} =\lim_{(x,m x)\to (0,0)}\frac{m^2x^3}{x^2+m^4x^4}=\lim_{x\to 0}\frac{m^2x}{1+m^4x^2}=0 . \end{equation} But along the curve $x=y^2$ we have \begin{equation} \lim_{ \begin{array}{c} (x,y)\to (0,0) \\ x=y^2 \end{array} }\dfrac{x y^2}{x^2+y^4} =\lim_{\left(y^2,y\right)\to (0,0)}\frac{y^4}{y^4+y^4}=\lim_{y\to 0}\frac{y^4}{y^4+y^4}=\frac{1}{2}. \end{equation} Therefore the given limit does not exist.

Example. Evaluate the following limit $\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{x^2y^2}{x^4+y^4}.$ A sketch of the function $f(x,y)=\dfrac{x^2y^2}{x^4+y^4}.$

Solution. Let $m$ be any real number, then along the paths $y=m x$ we have \begin{equation}
\lim_{ \begin{array}{c} (x,y)\to (0,0) \\ y=mx \end{array} }\dfrac{x^2y^2}{x^4+y^4} =\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^4+y^4} =\lim_{(x,m x)\to (0,0)}\frac{x^2(m x)^2}{x^4+(m x)^4} =\frac{m^2}{1+m^4}. \end{equation} Notice as the path $y=m x$ changes (as $m$ changes) so does the value $m^2/(1+m^4).$ Therefore the given limit does not exist.

Example. Evaluate the following limit $\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{x^4y^4}{\left(x^2+y^4\right)^3}.$ A sketch of the function $f(x,y)=\dfrac{x^4y^4}{\left(x^2+y^4\right)^3}.$

Solution. Along the line $y=0$ we have \begin{equation} \lim_{ \begin{array}{c} (x,y)\to (0,0) \\ y=0 \end{array} }\dfrac{x^4y^4}{\left(x^2+y^4\right)^3} =\lim_{x\to 0}\frac{0}{\left(x^2+0\right)^3}=0 \end{equation} and along the path $y=\sqrt{x}$ we have \begin{equation} \lim_{ \begin{array}{c} (x,y)\to (0,0) \\ y=\sqrt{x} \end{array} }\dfrac{x^4y^4}{\left(x^2+y^4\right)^3} =\lim_{x\to 0}\frac{x^4x^2}{\left(2x^2\right)^3}=\frac{1}{8}. \end{equation} Therefore the given limit does not exist.

Example. Find a function $f(x,y)$ and a point $\left(x_0,y_0\right)$ such that \begin{equation} \lim_{x\to x_0}\left(\lim_{y\to y_0}f(x,y)\right)\neq \lim_{y\to y_0}\left(\lim_{x\to x_0}f(x,y)\right). \end{equation}

Solution. Consider the function $f(x,y)=\frac{a x+b y}{c x+d y}$ and the point $(0,0).$ Then \begin{equation} \lim_{x\to 0}\left(\lim_{y\to 0}\frac{a x+b y}{c x+d y}\right)=\frac{a}{c}\neq \lim_{y\to 0}\left(\lim_{x\to 0}\frac{a x+b y}{c x+d y}\right)=\frac{b}{d}\end{equation} for some values of $a,b , c,$ and $d.$ What does this say about \begin{equation} \lim_{(x,y)\to (0,0)}\left(\frac{a x+b y}{c x+d y}\right) \end{equation}

Example. Consider the function $f(x,y)=\frac{x y }{x^2y^2+(x-y)^2}.$ Notice that \begin{equation} \label{multex} \lim_{x\to 0}\left(\lim_{y\to 0}f(x,y)\right)=\lim_{y\to 0}\left(\lim_{x\to 0}f(x,y)\right). \end{equation} Is it true, then, that \begin{equation} \label{dnemultex} \lim_{(x,y)\to (0,0)}\left(\frac{xy}{x^2y^2+(x-y)^2}\right) \end{equation} exists? A sketch of the function $f(x,y)= \frac{x y }{x^2y^2+(x-y)^2}.$

Solution. First we notice the (iterated) limits in \eqref{multex} are both $-1/2$ as the following shows \begin{align} & \lim_{x\to0}\left[ \lim_{y\to0} \left(\frac{xy}{x^2y^2+(x-y)^2}\right) \right] \\ & \qquad =\lim_{x\to0}\left[ \lim_{y\to0} \left(\frac{x}{2x^2y+2(x-y)(-1)}\right) \right] =-\frac{1}{2} \\ & \lim_{y\to0}\left[ \lim_{x\to0} \left(\frac{xy}{x^2y^2+(x-y)^2}\right) \right] \\ & \qquad =\lim_{y\to0}\left[ \lim_{x\to0} \left(\frac{y}{2y^2x+2(x-y)(1)}\right) \right] =-\frac{1}{2} \end{align} using L’Hopitals rule. Secondly, we find that the limit in \eqref{dnemultex} does not exist because along the paths $y=m x$ we have \begin{equation} \lim_{ \begin{array}{c} (x,y)\to (0,0) \\ y=mx \end{array} }\frac{xy}{x^2y^2+(x-y)^2} =\lim_{x\to 0}\left(\frac{mx^2}{m^2 x^4+(x-mx)^2}\right) =\frac{m}{(1-m)^2} \end{equation} which varies as the path $y=m x$ does.

## Continuity of a Function of Two Variables

Recall the definition of a continuous function from calculus 1.

Definition. Let $f$ be a function that is defined on any open disk around $(a,b).$ Then $f$ is continuous at the point $(a,b)$ if \begin{equation} \lim_{(x,y)\to (a,b)}f(x,y)=f(a,b). \end{equation}

Theorem. (Continuous Multivariable Functions) If $k$ is a real number and $f$ and both $g$ are continuous functions at $(a,b)$, then the following functions are also continuous at $(a,b)$, provided $(a,b)$ is in the domain of the function: $kf$, $f+g$, $f-g$, $f/g$, $f\circ g$, $\sqrt[n]{f}$

Proof. The basic properties of limits can be used to proof this theorem. The details are left for the reader.

Since any polynomial and rational function can be built out of the continuous functions $f(x)=x$, $g(x,y)=y$, and $h(x,y)=k$ we can use the limits in Continuous Multivariable Functions to realize that every polynomial is continuous on the entire plane and that rational functions are continuous on their domains.

Example. What is the largest set on which the function \begin{equation} f(x,y)= \begin{cases} \dfrac{3x^2y}{x^2+y^2} & \text{if} \quad (x,y)\neq (0,0) \\ 0& \text{if} \quad (x,y)=(0,0) \end{cases} \end{equation} is continuous? A sketch of the function $f$.

Solution. We know $f$ is continuous for $(x,y)\neq (0,0)$ since $f$ is a rational function. Since $$\lim_{(x,y)\to (0,0)}f(x,y)=0=f(0,0)$$ we have that $f$ is continuous on $\mathbb{R}^2.$

Example. What is the largest set on which the function \begin{equation} h(x,y)=\ln \left(x^2+y^2-1\right) \end{equation} is continuous? A sketch of the function $f(x,y)$.

Solution. Let $f(x,y)=x^2+y^2-1$ and $g(t)=\ln t.$ Then \begin{equation} h(x,y)=\ln \left(x^2+y^2-1\right)=(g\circ f)(x,y). \end{equation} Now $f$ is continuous everywhere since it is a polynomial and $g$ is continuous on its domain ${t \mid t\geq 0}.$ Thus, $h$ is continuous on its domain $$D=\left\{(x,y) \mid x^2+y^2>1 \right\}$$ which consists of all points outside the circle $x^2+y^2=1.$

## Exercises Continuous Functions and Multivariable Limits

Exercise. Where is $f(x,y)=\sqrt{\dfrac{x^2+y^2}{(x-y)^2+1}}$ is continuous?

Exercise. Given $\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{x^4y^4}{\left(x^2y^2\right)^{3/2}}$, find the limit.

Exercise. Find $\displaystyle \lim_{(x,y)\to (\pi /2,1)}\dfrac{1+\cos 2x}{y-e^y}.$
Explain why.

Exercise. Evaluate the limit.

$(1) \quad$ $\displaystyle \lim_{(x,y)\to (-1,0)}\left(x y^2+x^3y+5\right)$

$(2) \quad$ $\displaystyle \lim_{(x,y)\to (0,0)}\left(5x^2-2 x y+y^2+3\right)$

$(3) \quad$ $\displaystyle \lim_{(x,y)\to (0,1)}e^{x^2+x}\ln \left(e y^2\right)$

$(4) \quad$ $\displaystyle \lim_{(x,y)\to (a,a)}\frac{x^4-y^4}{x^2-y^2}$

$(5) \quad$ $\displaystyle \lim_{(x,y)\to (1,2)}\frac{\left(x^2-1\right)\left(y^2-4\right)}{(x-1)(y-2)}$

$(6) \quad$ $\displaystyle \lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^4+y^4}$

$(7) \quad$ $\displaystyle \lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$

$(8) \quad$ $\displaystyle \lim_{(x,y)\to (0,0)}\sin x+ \sin y$

$(9) \quad$ $\displaystyle \lim_{(x,y)\to (0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+4}-2}$

$(10) \quad$ $\displaystyle \lim_{(x,y)\to (0,0)}\frac{1-\cos \left(x^2+y^2\right)}{x^2+y^2}$

$(11) \quad$ $\displaystyle \lim_{(x,y)\to (0,0)}\frac{\sin \left(x^3+y^3\right)}{\sqrt{x^6+y^6}}$

$(12) \quad$ $\displaystyle \lim_{(x,y)\to (-1,0)}\left(x y^2+x^3y+5\right)$

$(13) \quad$ $\displaystyle \lim_{(x,y)\to (0,0)}\left(5x^2-2 x y+y^2+3\right)$

$(14) \quad$ $\displaystyle \lim_{(x,y)\to (0,1)}e^{x^2+x}\ln \left(e y^2\right)$

$(15) \quad$ $\displaystyle \lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$

$(16) \quad$ $\displaystyle \lim_{(x,y)\to (0,0)}\sin x+ \sin y$

$(17) \quad$ $\displaystyle \lim_{(x,y)\to (0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+4}-2}$

$(18) \quad$ $\displaystyle \lim_{(x,y)\to (0,0)}\frac{1-\cos \left(x^2+y^2\right)}{x^2+y^2}$

$(19) \quad$ $\displaystyle \lim_{(x,y)\to (0,0)}\frac{\sin \left(x^3+y^3\right)}{\sqrt{x^6+y^6}}$

Exercise. Is the function $f$ defined by \begin{equation} f(x,y)= \begin{cases} \dfrac{x y^2}{x^2+y^4} & \quad (x,y)\neq (0,0) \\ 0 & \quad (x,y)=(0,0) \end{cases} \end{equation} continuous at $(0,0)?$ Explain.

Exercise. Is the function $g$ defined by \begin{equation} g(x,y)= \begin{cases}
\dfrac{x y^3}{x^2+y^6} & \quad (x,y)\neq (0,0) \\ 0 & \quad (x,y)=(0,0) \end{cases} \end{equation} continuous at $(0,0)?$ Explain.

Exercise. Given that the function $f$ defined by \begin{equation} f(x,y)= \begin{cases} \dfrac{3x^3-3y^3}{x^2-y^2} & \quad x^2\neq y^2 \\ B & \text{otherwise} \end{cases} \end{equation} is continuous at the point $(0,0)$ what is the value of $B?$ Explain.

Exercise. Given that the function $f$ defined by \begin{equation}f(x,y)=\begin{cases}
\dfrac{x^3+y^3}{x^2+y^2} & \quad (x,y)\neq (0,0) \\ A & \quad (x,y)=(0,0) \end{cases} \end{equation} is continuous at the point $(0,0)$ what is the value of $A?$ Explain.

Exercise. Let $$f(x,y)= \begin{cases} \dfrac{x^2 y^2}{x^4+y^4+1} & \text{if } (x,y)\neq (0,0) \\ 0 & \text{if } (x,y)= (0,0). \end{cases}$$ Explain why $f$ is continuous at $(0,0).$