## 2.1 Multivariable Functions

Partial differentiation is a powerful mathematical tool that can be used to solve a variety of problems. In this book, we will discuss the fundamental concepts and principles of partial differentiation. We will also provide some examples to illustrate how partial differentiation can be used to find maxima and minima, optimize functions, and more.

Partial differentiation is the process of taking the partial derivative of a function with respect to one or more of its variables. The partial derivative of a function is the rate of change of that function with respect to one of its variables, holding all other variables constant. Partial differentiation is used in calculus and other mathematical disciplines to find maxima and minima, as well as to study the behavior of functions near points of discontinuity.

Partial differentiation has two key concepts: the derivative and the limit. The derivative is a measure of how a function changes as one of its inputs changes. The limit is the value that a function approaches as one of its inputs approaches a particular value. Partial differentiation relies on both of these concepts to quantify how a function changes near a point in space. In order to take the partial derivative of a function, we must first be able to take the derivative with respect to one variable while holding all other variables constant.

Partial differentiation is a powerful tool that can be used to solve problems in many different fields, including physics, engineering, and economics.

A multivariable function is a mathematical function of two or more variables. Examples of multivariable functions include the position of a moving particle in space (which depends on its velocity and acceleration), the electric and magnetic fields in electromagnetism, and the pressure, temperature, and density of a fluid in fluid mechanics. Multivariable functions can be graphed in three-dimensional space, and their behavior can be studied using calculus. In general, multivariable functions are more complicated than their single-variable counterparts. However, they can often be decomposed into simpler functions, making them easier to understand and work with.

Multivariable functions can be partial differentiated with respect to each of their variables. This partial differentiation is a measure of how the function changes when one of its variables is changed, while the other variables are held constant. In other words, it helps us to understand how a function behaves when we change one of its inputs. Partial differentiation can be used to optimize a function, by finding the input that will produce the desired output. It can also be used to find maxima and minima, and to solve problems in physics and engineering. So, if you’re ever stuck trying to figure out a multivariable function, remember that partial differentiation is your friend.

Multivariable limits and continuity are often some of the most challenging concepts for students to grasp in mathematics. In partial differentiation, we take the limit of a function as one or more of its variables approach infinity while the others remain finite. This can be a difficult concept to wrap one’s head around, but it is crucial for understanding many advanced mathematical concepts.

Continuity, on the other hand, is a relatively straightforward concept: it simply means that a function is uninterrupted and smooth. However, proving continuity can often be quite tricky. In general, multivariable limits and continuity can be quite challenging concepts, but they are essential for understanding advanced mathematics.

When it comes to math, there are a lot of big words that can be intimidating. Differentials and differentiability may sound like they belong in a horror movie, but they’re actually just two concepts that are closely related.

In calculus, a differential is a small change in a variable, while differentiability refers to the ability to calculate these small changes. These concepts are important in calculus and other branches of mathematics, as they allow for the determination of things like rates of change and slopes of curves. So the next time you see a differential or differentiability, don’t be scared off - they’re just math terms!

The chain rule is one of the most important concepts in calculus, and it can be a little tricky to wrap your head around at first. Basically, the chain rule says that if you have a function that is the composition of two other functions, you can take the derivative of that function by taking the derivative of each of the individual functions and then multiplying them together.

The chain rule can be generalized to functions with more than two components, but the basic idea is always the same: to take the derivative of a composite function, you need to take the derivative of each individual function and then multiply them all together.

Partial differentiation can also be used when taking derivatives with respect to more than one variable. However, it’s important to remember that this rule is only useful with functions that are composed of other functions - if a function is not composed of other functions, then the chain rule is not that useful at all. So, if you ever come across a function that you’re not sure how to differentiate, make sure to check whether or not it’s composed of other functions before trying using the chain rule.

Partial differentiation simply means taking the derivative with respect to one variable while holding all other variables constant. For example, consider the function $$f(x,y) = x^2 + y^2$$. The partial derivative of this function with respect to x is 2x, and the partial derivative with respect to y is 2y. This means that as we move across the graph of this function, it is increasing at a rate of 2x in the direction of x and at a rate of 2y in the direction of y.

The gradient vector is simply a vector that contains all partial derivatives of a given function. So, for our example function $$f(x,y) = x^2 + y^2$$, the gradient vector would be $$\langle2x, 2y\rangle$$. This vector can be thought of as a directional derivative” because it tells us which way the function is increasing and by how much. The magnitude of the gradient vector tells us how steeply the function is increasing (i.e., how steeply it is “pointing” in a particular direction), and the direction tells us which way it is pointing.

There are many applications for partial derivatives and gradient vectors in physics and engineering. One common application is finding local maxima and minima (i.e., points where the derivative is zero).

Another common application is optimization; for example, finding the shortest path between two points or minimizing fuel consumption in a car engine. Partial derivatives and gradient vectors also play a central role in Newton’s Method for solving differential equations numerically.

In short, partial derivatives and gradient vectors are essential tools for anyone who wants to understand how functions change as their inputs vary.

Differentiation is all about finding rates of change. And in the world of partial differential equations, we’re usually interested in rates of change with respect to more than one variable. Enter the tangent plane and the normal line.

The tangent plane to a surface at a point is the plane that just barely touches the surface at that point. It’s like if you took a really thin piece of paper and tried to flatten it out on the surface. The normal line is the line that is perpendicular to the tangent plane at the point of contact.

Partial differentiation is all about taking derivatives with respect to more than one variable, so tangent planes and normal lines are going to be our best friends. We can use them to find rates of change in multiple directions, which will come in handy when we’re trying to solve partial differential equations.

Partial differentiation is a way of finding the extreme values of a function of two variables. In other words, it helps us find the points where the function is at its highest or lowest.

To do this, we take partial derivatives with respect to each of the variables. The partial derivative with respect to x tells us how the function changes as x changes, and the partial derivative with respect to y tells us how the function changes as y changes. If both partial derivatives are positive, then the function is increasing in both variables, and so the point is not an extreme value.

However, if both partial derivatives are negative, then the function is decreasing in both variables, and so the point is an extreme value. If one partial derivative is positive and the other is negative, then the function is stylized and we need to use the second partial derivative test to determine whether the point is an extreme value.

So, partial differentiation can be used to find extreme values of functions of two variables.

What are Lagrange multipliers? In short, they’re a mathematical way of solving optimization problems. But what does that mean, exactly?

Partial differentiation is a powerful tool that allows us to optimize functions subject to constraints. In many cases, the constrained optimization problem can be simplified by introducing Lagrange multipliers. Lagrange multiplier methods are named after Joseph-Louis Lagrange, who first developed the technique in the 18th century.

The key idea is to introduce new variables, called Lagrange multipliers, which represent the constraints of the optimization problem. These multipliers can then be used to simplify the optimization problem by eliminating the need for explicit constraints. As a result, the Lagrange multiplier method is a powerful tool for solving constrained optimization problems.

If you’re interested in learning partial differentiation, this book is for you. It covers all the basics, from the definition of partial derivatives to the chain rule and beyond. Plus, it includes plenty of worked examples to help you understand and master every concept.

How does one go about teaching partial differentiation? It’s simply a matter of conveying an understanding clearly and concisely. Of course, that’s often easier said than done. partial differentiation can be a tricky topic to wrap your head around, let alone explain to someone else. But with a little patience and perseverance, you should be able to do it. After all, if you can partially differentiate, anyone can!

## 2.2 Functions of Several Variables

A polynomial function of two variables is a sum of terms of the form $$cx^m y^n$$, where $$c$$ is a real number and both $$m$$ and $$n$$ are nonnegative integers. For example the function $f(x,y)=3x^3 y^5-3x^2 y^4-x y +7$ is a polynomial function with terms $$3x^3y^5$$, $$-3x^2 y^4$$, $$-xy$$, and $$7$$. A rational function of two variables is a ratio of polynomial functions. For example the function $f(x,y)=\frac{3x^3 y^5-3x^2 y^4-x y +7}{x^3-y^2}$ is a rational function with numerator $$3x^3 y^5-3x^2 y^4-x y +7$$ and denominator $$x^3-y^2$$.

A function of three variables is a rule that assigns to each ordered pair $$\left(x, y, z\right)$$ in a set $$D$$ a unique number $$f\left(x, y, z\right).$$ The set $$D$$ is called the domain of the function and is a subset of $$\mathbb{R}^3$$. The collection of corresponding values $$f\left(x,y,z\right)$$ constitutes the range which is a subset of $$\mathbb{R}$$.

Example 2.1 Let $f(x,y,z)=x^2-2xy z+2 y z-x.$ Find the domain of $$f$$ and evaluate $$f(1,2,3)$$, $$f(t,2t,3t)$$, $$f(x+y, x-y, 0)$$, and $\frac{f(x+h,y,z)-f(x,y,z)}{h}, \quad \frac{f(x,y+h,z)-f(x,y,z)}{h}, \quad \frac{f(x,y,z+h)-f(x,y,z)}{h}.$ where $$h$$ is an unknown quantity.

Solution. Substitution into the function yields the following: $$f(1,2,3)=0$$, $f(t,2t,3t)=-t+13 t^2-12 t^3 \qquad f(x+y,x-y,0)=-x-y+(x+y)^2$ and \begin{align} & \frac{f(x+h,y,z)-f(x,y,z)}{h} = \frac{-h-x^2+(h+x)^2+2 x y z-2 (h+x) y z}{h} \\ & \frac{f(x,y+h,z)-f(x,y,z)}{h} = \frac{-h-x^2+(h+x)^2+2 x y z-2 (h+x) y z}{h} \\ & \frac{f(x,y,z+h)-f(x,y,z)}{h} = \frac{2 y z-2 x y z-2 y (h+z)+2 x y (h+z)}{h} \end{align} The domain of $$f$$ is all real numbers, but rather $$\mathbb{R}^3$$.

Definition 2.1 A function of $$n$$ variables is a rule that assigns to each ordered pair $$\left(x_1, x_2, \ldots , x_n\right)$$ in a set $$D$$ a unique number $$f\left(x_1, x_2, \ldots , x_n\right).$$ The set $$D$$ is called the domain of the function and the set of corresponding values $$f\left(x_1, x_2, \ldots , x_n\right)$$ is called the range.

The graph of a function of several variables $$f\left(x_1,\ldots,x_n\right)$$ is the collection of all ordered $$(n+1)$$-tuples $$\left(x_1,\ldots,x_n,x_{n+1}\right)$$ such that $$\left(x_1,\ldots,x_n\right)$$ is in the domain of $$f$$ and $$x_{n+1}=f\left(x_1,\ldots,x_n\right).$$ Sketching by hand the graph of a function with several variables can be challenging. Let’s look at a few particular functions.

For example the function $$f$$ defined by $$$\label{polyfunc} f(x,y,z)=2x^3+4y^4+9z^6$$$ is a three variable polynomial function with domain $$\mathbb{R}^3$$ and range $$(-\infty, +\infty)$$.
As another example consider the function $$g$$ defined by $$$\label{trigfunc} g(w,x,y,z)=z^2+\sin w x+\cos w y.$$$ The function $$g$$ is a four variable function with domain $$\mathbb{R}^4$$ and range $$[-2,+\infty)$$. Let $$h$$ be the function defined by $$$\label{expfunc} h(x,y,z)=e^{xy}+z^4\sqrt[6]{x^2-4}.$$$ Then $$h$$ is a three variable function with domain all ordered triples $$(x,y,z)$$ with the requirement $$x^2-4\geq 0$$. The range of $$h$$ is the set of all real numbers greater than 0.

Definition 2.2 Let $$f$$ and $$g$$ be functions of the variables $$x_1, x_2, \ldots,x_n$$. Then defined point-wise, the following functions

• $$(f+g)\left(x_1, x_2, \ldots, x_n\right):= f\left(x_1, x_2, \ldots, x_n\right)+g\left(x_1, x_2, \ldots, x_n\right)$$
• $$(f-g)\left(x_1, x_2, \ldots, x_n\right):= f\left(x_1, x_2, \ldots, x_n\right)-g\left(x_1, x_2, \ldots, x_n\right)$$
• $$(f g)\left(x_1, x_2, \ldots, x_n\right):= f\left(x_1, x_2, \ldots, x_n\right)g\left(x_1, x_2, \ldots, x_n\right)$$
• $$(f/g)\left(x_1, x_2, \ldots, x_n\right):=\dfrac{f\left(x_1, x_2, \ldots, x_n \right)}{g\left(x_1, x_2, \ldots, x_n\right)}$$ where $$g\left(x_1, x_2, \ldots, x_n\right)\neq 0.$$

are also functions of the variables $$x_1, x_2, \ldots, x_n$$.

For example, considering the functions $$f$$ and $$h$$ defined in $$\ref{polyfunc}$$ and $$\ref{expfunc}$$, respectively. Is $$f+h$$ defined as a function? Yes, by $$\ref{ptwisedeffunc}$$
$(f+h)(x,y,z)= 2x^3+4y^4+9z^6+e^{xy}+z^4\sqrt[6]{x^2-4}$ and notice the domain of $$f+h$$ is all ordered triples $$(x,y,z)$$ with the requirement $$x^2-4\geq 0$$ and the range is $$(-\infty,+\infty)$$.

Is $$g+h$$ defined as a function? If we use the functional rule defining $$h$$, namely $$e^{xy}+z^4\sqrt[6]{x^2-4}$$ then the function defined by $h_1(w,x,y,z)=e^{xy}+z^4\sqrt[6]{x^2-4}$ (defined as a four variable function), then the function $$(g+h_1)$$ is defined as a function by $(g+h_1)(w,x,y,z)=z^2+\sin w x+\cos w y+ e^{xy}+z^4\sqrt[6]{x^2-4}.$ using $$\ref{ptwisedeffunc}$$. Notice the domain of $$g+h_1$$ is $$\mathbb{R}^4$$ and the range is $$[-2,+\infty)$$.

## 2.3 Functions of Two Variables

When working with functions $$f$$ of two variables $$x$$ and $$y$$, we write $$z=f(x,y)$$ where $$x$$ and $$y$$ are the independent variables and $$z$$ is the dependent variable. The domain is defined to be the largest set of points for which the functional formula is defined and real-valued.

Example 2.2 Find the domain and range for the function $f(x,y)=\dfrac{1}{\sqrt{x-y}}.$

Solution. The domain is $\{(x,y)\in \mathbb{R}^2 \mid y<x\}$ because of the square root in the denominator of $$f$$ and the range is $\{z \in \mathbb{R} \, \mid \, z>0\}.$ Notice the domain is a subset of $$\mathbb{R}^2$$ and the range is a subset of $$\mathbb{R}$$.

Example 2.3 Find the domain and range for the function $f(x,y)=\sqrt{\dfrac{x}{y}}.$

Solution. The domain is $\{(x,y)\in \mathbb{R}^2\, \mid\, x y\geq 0 \text{ and } y\neq 0\}$ because of the square root and the range is $\{z \in \mathbb{R} \, \mid \, z\geq 0\}.$ Notice the domain is a subset of $$\mathbb{R}^2$$ and the range is a subset of $$\mathbb{R}$$.

In three dimensions, the graph of $$z=f(x,y)$$ is a surface in $$\mathbb{R}^3$$ whose projection onto the $$x y$$-plane is the domain $$D$$. When the plane $$z=C$$ intersects the surface $z=f(x,y),$ the result is the curve with the equation $$f(x,y)=C$$ and such an intersection is called the trace of the graph of $$f$$ in the plane $$z=C.$$ The set of points $$(x,y)$$ in the $$x y$$-plane that satisfies $f(x,y)=C$ is called the level curve of $$f$$ at $$C$$ and an entire family of level curves is generated as $$C$$ varies over the range of $$f$$. Level curves are obtained by projecting a trace onto the $$x y$$-plane. Because level curves are used to show the shape of a surface, they are sometimes called contour curves .

Definition 2.3 The curves $$f(x,y)=C$$ in the $$xy$$-plane are called the level curves of the function $$f$$ of two variables $$x$$ and $$y$$, where $$C$$ is a constant in the range of $$f$$.

Example 2.4 Sketch a few level curves for the function $f(x,y)=2x-3y=C$ with $$C\geq 0$$.

Solution. Graphing the lines $y=\frac{2}{3}x-\frac{C}{3}$ we have the family of level curves corresponding to $$C=1,\ldots,10.$$ The level curves show that the graph of the function $$f$$ is a plane in $$\mathbb{R}^3$$.

Example 2.5 Sketch a few level curves for the function $f(x,y)=x^2+\frac{y^2}{4}=C$ with $$C\geq 0$$.

Solution. Graphing the ellipses $\frac{x^2}{1^2}+\frac{y^2}{2^2}=C$ in $$\mathbb{R}^2,$$ we have the family of level curves corresponding to $$C=1,\ldots,10.$$ The level curves show that the graph of the function $$f$$ is a elliptic paraboloid in $$\mathbb{R}^3$$.

Example 2.6 Find the domain and range for the function $f(x,y)=\sqrt{\frac{y}{x-2}}$ and sketch some level curves for $$f(x,y)=C$$ with $$C=0,1,2,3,4$$.

Solution. The domain of $$f$$ is $\left\{(x,y)\in \mathbb{R}^2 \mid \frac{y}{x-2}>0 \right\}.$ To sketch some level curves $\sqrt{\frac{y}{x-2}}=C$ let’s square both sides $$\frac{y}{x-2}=C^2$$ and so $y=C^2(x-2).$ The range of $$f$$ is $$\{z\in \mathbb{R} \mid z\geq 0\}$$.

## 2.4 Exercises

Exercise 2.1 Let $$f(x,y,z)=x^2y e^{2x}+(x+y-z)^2.$$ Find each of the following.

• $$f(0,0,0)$$
• $$f(1,-1,1)$$
• $$f(-1,1,-1)$$
• $$\frac{d}{dx}f(x,x,x),$$
• $$\frac{d}{dy}f(1,y,1)$$
• $$\frac{d}{dz}f\left(1,1,z^2\right)$$

Exercise 2.2 Find the domain and range for the multivariate function.

• $$f(x,y)=\frac{1}{\sqrt{x-y}}$$
• $$f(x,y)=\sqrt{\frac{y}{x}}.$$
• $$f(u,v)=\sqrt{u \sin v}.$$
• $$f(x,y)=e^{(x+1)/(y-2)}.$$
• $$f(x,y)=\frac{1}{\sqrt{9-x^2-y^2}}.$$

Exercise 2.3 Sketch some level curves of the function.

• $$f(x,y)=x^2-y^2=C$$
• $$f(x,y)=\frac{x}{y}=C$$
• $$f(x,y)=x^2-y=C$$
• $$f(x,y)=x^2+\frac{y^2}{4}=C$$
• $$f(x,y)=x^3-y=C$$
• $$f(x,y)=1/\sqrt{x^2-y^2}$$
• $$g(x,y)=\sqrt{x \sin y}$$
• $$h(x,y)=\ln (y-x)$$

Exercise 2.4 Describe the trace of the quadratic surface in each coordinate plane, then sketch the surface.

• $$\frac{x^2}{4}+y^2+\frac{z^2}{9}=1$$
• $$\frac{x^2}{9}-y^2-z^2=1$$

Exercise 2.5 Sketch the graph of the multivariate function.

• $$f(x,y)=x$$
• $$f(x,y)=x^3-1$$
• $$f(x,y)=x^2-y$$
• $$f(x,y)=x^2-y^2$$
• $$f(x,y)=\sqrt{x+y}$$
• $$f(x,y)=1/\sqrt{x^2-y^2}$$
• $$g(x,y)=\sqrt{x \sin y}$$
• $$h(x,y)=\ln (y-x)$$

## 2.5 Multivariable Limits

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 $\sqrt{(x-a)^2+(y-b)^2}<r$ for $$r>0$$. If the boundary of the disk is included, the disk is called a closed disk.

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 2.4 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 $\lim_{(x,y)\to(a,b)} f(x,y)=L$ if $$f(x,y)$$ can be made as close to $$L$$ as we please by restricting $$(x,y)$$ to be sufficiently close to $$(a,b)$$.

## 2.6 Limit Properties

If $$a$$, $$b$$, and $$c$$ are real numbers then $$$\label{linmultlimit} \lim_{(x,y)\to(a,b)}c=c,\qquad \lim_{(x,y)\to(a,b)}x=a, \qquad \lim_{(x,y)\to(a,b)}y=b.$$$ The three limits in $$\ref{linmultlimit}$$ can be used with $$\ref{limthmmult}$$ to show that $$$\label{polratlimit} \lim_{(x,y)\to(a,b)}p(x,y)=p(a,b) \qquad \text{and}\qquad \lim_{(x,y)\to(a,b)}r(x,y)=r(a,b)$$$ 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 $$r$$.

::: {#thm- } [Limit Law] 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

• $$\lim _{(x,y)\to \left(x_0,y_0\right)}(a f)(x,y)=a L$$
• $$\lim _{(x,y)\to \left(x_0,y_0\right)}(f+g)(x,y)=L+M$$
• $$\lim _{(x,y)\to \left(x_0,y_0\right)}(f g)(x,y)=L M$$
• $$\lim _{(x,y)\to \left(x_0,y_0\right)}(f/g)(x,y)=L/M$$
whenever $$M\neq 0.$$ :::

Proof. The proof is left for the reader.

Example 2.7 Evaluate $$\lim _{(x,y)\to (1,1)}\dfrac{x^4+y^4}{x^2+y^2}$$.

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 use $$\ref{limthmmult}$$ to find $\lim _{(x,y)\to (1,1)}\dfrac{x^4+y^4}{x^2+y^2} =r(2,2)=1.$

Example 2.8 Evaluate $$\lim _{(x,y)\to (0,0)}\dfrac{x^4-y^4}{x^2+y^2}$$.

Solution. By canceling $$$\label{cancellim} \lim _{(x,y)\to (0,0)}\frac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{x^2+y^2}=\lim _{(x,y)\to (0,0)}\left(x^2-y^2\right)=0.$$$ Notice the functions $$f$$ and $$g$$ defined by $f(x,y)=x^2-y^2 \qquad \text{and}\qquad g(x,y)=\dfrac{x^4-y^4}{x^2+y^2}$ 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 $$\ref{cancellim}$$ is valid.

Example 2.9 Evaluate $$\lim_{(x,y)\to (1,2)}\dfrac{\left(x^2-1\right)\left(y^2-4\right)}{(x-1)(y-2)}$$.

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. \tag*{} \end{align*}

Example 2.10 Evaluate $$\lim _{(x,y)\to (a,a)}\dfrac{x^4-y^4}{x^2-y^2}$$.

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. \tag*{} \end{align*}

## 2.7 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.

::: {#thm- } [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. :::

Proof. The proof is left for the reader.

Example 2.11 Evaluate $$\lim _{(x,y)\to (0,0)}\dfrac{x y}{x^2+y^2}$$.

Solution. Along the $$x$$-axis we have $\plim{(x,y)\to (0,0)}{y=0}\dfrac{x y}{x^2+y^2} =\lim _{(x,0)\to (0,0)}\frac{0}{x^2+0^2}=\lim _{x\to 0}0=0.$ We have along the $$y$$-axis, $\plim{(x,y)\to (0,0)}{x=0}\dfrac{x y}{x^2+y^2} =\lim _{(0,y)\to (0,0)}\frac{0}{0^2+y^2}=\lim _{y\to 0}0=0.$ Along the line $$y=x$$ we have $\plim{(x,y)\to (0,0)}{y=x}\dfrac{x y}{x^2+y^2} =\lim _{(x,x)\to (0,0)}\frac{ x^2 }{x^2+x^2}=\lim _{x\to 0}\frac{1}{2}=\frac{1}{2}.$
Therefore, by $$\ref{dnelim}$$, the given limit does not exist (see $$\ref{fig:limdne}$$).

Example 2.12 Evaluate $$\lim _{(x,y)\to (0,0)}\dfrac{x y^2}{x^2+y^4}$$.

Solution. Along the curves $$y=m x$$ we have $\plim{(x,y)\to (0,0)}{y=mx}\dfrac{x y^2}{x^2+y^4} =\lim _{(x,m x)\to (0,0)}\frac{m^2x^3}{x^2+m^4x^4}=\lim _{x\to 0}\frac{m^2x}{1+m^4x^2}=0 .$ But along the curve $$x=y^2$$ we have $\plim{(x,y)\to (0,0)}{x=y^2}\dfrac{x y^2}{x^2+y^4} =\lim _{\left(y^2,y\right)\to (0,0)}\frac{y^4}{y^4+y^4}=\lim _{y\to 0}\frac{y^4}{y^4+y^4}=\frac{1}{2}.$ Therefore, by $$\ref{dnelim}$$, the given limit does not exist.

Example 2.13 Evaluate $$\lim _{(x,y)\to (0,0)}\dfrac{x^2y^2}{x^4+y^4}$$.

Solution. Let $$m$$ be any real number, then along the paths $$y=m x$$ we have $\plim{(x,y)\to (0,0)}{y=mx}\dfrac{x^2y^2}{x^4+y^4} =\lim _{(x,y)\to (0,0)}\frac{x^2y^2}{x^4+y^4} =\lim _{(x,m x)\to (0,0)}\frac{x^2(m x)^2}{x^4+(m x)^4} =\frac{m^2}{1+m^4}.$ Notice as the path $$y=m x$$ changes (as $$m$$ changes) so does the value $$m^2/(1+m^4)$$. Therefore, by $$\ref{dnelim}$$, the given limit does not exist.

Example 2.14 Evaluate $$\lim _{(x,y)\to (0,0)}\dfrac{x^4y^4}{\left(x^2+y^4\right)^3}$$.

Solution. Along the line $$y=0$$ we have $\plim{(x,y)\to (0,0)}{y=0}\dfrac{x^4y^4}{\left(x^2+y^4\right)^3} =\lim _{x\to 0}\frac{0}{\left(x^2+0\right)^3}=0$ and along the path $$y=\sqrt{x}$$ we have $\plim{(x,y)\to (0,0)}{y=\sqrt{x}}\dfrac{x^4y^4}{\left(x^2+y^4\right)^3} =\lim_{x\to 0}\frac{x^4x^2}{\left(2x^2\right)^3}=\frac{1}{8}.$ Therefore, by $$\ref{dnelim}$$, the given limit does not exist.

Example 2.15 Find a function $$f(x,y)$$ and a point $$\left(x_0,y_0\right)$$ such that $\lim _{x\to x_0}\left(\lim _{y\to y_0}f(x,y)\right)\neq \lim _{y\to y_0}\left(\lim _{x\to x_0}f(x,y)\right).$

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

Example 2.16 Consider the function $f(x,y)=\frac{x y }{x^2y^2+(x-y)^2}.$ Notice that
$$$\label{multex} \lim _{x\to 0}\left(\lim _{y\to 0}f(x,y)\right)=\lim _{y\to 0}\left(\lim _{x\to 0}f(x,y)\right).$$$ Is it true, then, that $$$\label{dnemultex} \lim _{(x,y)\to (0,0)}\left(\frac{xy}{x^2y^2+(x-y)^2}\right)$$$ exists?

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

## 2.8 Continuity of a Function of Two Variables

Definition 2.5 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 $\lim _{(x,y)\to (a,b)}f(x,y)=f(a,b).$

::: {#thm- } [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 $$\ref{multlimprop}$$ to realize that every polynomial is continuous on the entire plane and that rational functions are continuous on their domains.

Example 2.17 What is the largest set on which the function $f(x,y)=\left\{ \begin{array}{ll} \dfrac{3x^2y}{x^2+y^2} & \text{if} \quad (x,y)\neq (0,0) \\ 0& \text{if} \quad (x,y)=(0,0) \end{array} \right.$ is continuous?

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 2.18 What is the largest set on which the function $h(x,y)=\ln \left(x^2+y^2-1\right)$ is continuous?

Solution. Let $$f(x,y)=x^2+y^2-1$$ and $$g(t)=\ln t$$. Then $h(x,y)=\ln \left(x^2+y^2-1\right)=(g\circ f)(x,y).$ 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 $$x^2+y^2=1.$$

## 2.9 Exercises

Exercise 2.6 Where is $$f(x,y)=\sqrt{\frac{x^2+y^2}{(x-y)^2+1}}$$ is continuous?

Exercise 2.7 Given $$\lim _{(x,y)\to (0,0)}\frac{x^4y^4}{\left(x^2y^2\right)^{3/2}}$$, find the limit.

Exercise 2.8 Find $$\lim _{(x,y)\to (\pi /2,1)}\frac{1+\cos 2x}{y-e^y}.$$ Explain why.

Exercise 2.9 Evaluate the limit.

• $$\lim_{(x,y)\to (-1,0)}\left(x y^2+x^3y+5\right)$$
• $$\lim_{(x,y)\to (0,0)}\left(5x^2-2 x y+y^2+3\right)$$
• $$\lim_{(x,y)\to (0,1)}e^{x^2+x}\ln \left(e y^2\right)$$
• $$\lim_{(x,y)\to (a,a)}\frac{x^4-y^4}{x^2-y^2}$$
• $$\lim_{(x,y)\to (1,2)}\frac{\left(x^2-1\right)\left(y^2-4\right)}{(x-1)(y-2)}$$
• $$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^4+y^4}$$
• $$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$$
• $$\lim_{(x,y)\to (0,0)}\sin x+ \sin y$$
• $$\lim_{(x,y)\to (0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+4}-2}$$
• $$\lim_{(x,y)\to (0,0)}\frac{1-\cos \left(x^2+y^2\right)}{x^2+y^2}$$
• $$\lim_{(x,y)\to (0,0)}\frac{\sin \left(x^3+y^3\right)}{\sqrt{x^6+y^6}}$$
• $$\lim_{(x,y)\to (-1,0)}\left(x y^2+x^3y+5\right)$$
• $$\lim_{(x,y)\to (0,0)}\left(5x^2-2 x y+y^2+3\right)$$
• $$\lim_{(x,y)\to (0,1)}e^{x^2+x}\ln \left(e y^2\right)$$
• $$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$$
• $$\lim_{(x,y)\to (0,0)}\sin x+ \sin y$$
• $$\lim_{(x,y)\to (0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+4}-2}$$
• $$\lim_{(x,y)\to (0,0)}\frac{1-\cos \left(x^2+y^2\right)}{x^2+y^2}$$
• $$\lim_{(x,y)\to (0,0)}\frac{\sin \left(x^3+y^3\right)}{\sqrt{x^6+y^6}}$$

Exercise 2.10 Is the function $$f$$ defined by $f(x,y)= \begin{cases} \dfrac{x y^2}{x^2+y^4} & \quad (x,y)\neq (0,0) \\ 0 & \quad (x,y)=(0,0) \end{cases}$ continuous at $$(0,0)?$$ Explain.

Exercise 2.11 Is the function $$f$$ defined by $g(x,y)= \begin{cases} \dfrac{x y^3}{x^2+y^6} & \quad (x,y)\neq (0,0) \\ 0 & \quad (x,y)=(0,0) \end{cases}$ continuous at $$(0,0)?$$ Explain.

Exercise 2.12 Given that the function $$f$$ defined by $f(x,y)= \begin{cases} \dfrac{3x^3-3y^3}{x^2-y^2} & \quad x^2\neq y^2 \\ B & \text{otherwise} \end{cases}$ is continuous at the point $$(0,0)$$ what is the value of $$B?$$ Explain.

Exercise 2.13 Given that the function $$f$$ defined by $f(x,y)= \begin{cases} \dfrac{x^3+y^3}{x^2+y^2} & \quad (x,y)\neq (0,0) \\ A & \quad (x,y)=(0,0) \end{cases}$ is continuous at the point $$(0,0)$$ what is the value of $$A?$$ Explain.

Exercise 2.14 Let $$f(x,y)=\begin{cases} \frac{x^2 y^2}{x^4+y^4+1} & \text{if } (x,y)\neq (0,0) \\ 0 & \text{if } (x,y)= (0,0). \end{cases}$$ Explain why $$f$$ is continuous at $$(0,0)$$.