Math 344: Calculus III

14.2 Limits and Continuity

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14.2.1 Definition of a Limit for Multivariate Functions

For functions of two (or more) variables, it is necessary to build up the notions of limits and continuity before we can discuss the details of how to use calculus with these functions. Fortunately, the definition of limits (and thus continuity) do not differ much from the one we are used to working with. It still follows the $\epsilon$-$\delta$ structure with small modifications to the domain and codomain involved and how distance is accounted for.

Definition

Limit of a Multivariate Function
Let $f$ be a function of two (or more) variables whose domain $D\subset\mathbb{R}^n$ includes points in the neighborhood of a point $\mathbf{a}\in\mathbb{R}^n$ (that is all points in an arbitrarily small open set including $\mathbf{a}$). Then the limit of $f(\mathbf{x})$ as $\mathbf{x}$ approaches $\mathbf{a}$ is $L$ and is denoted by

$$ \lim_{\mathbf{x}\rightarrow \mathbf{a}} f(\mathbf{x}) = L $$
if for each error bound $\epsilon\gt 0$ there is a corresponding input precision $\delta\gt 0$ such that if $\mathbf{x}\in D$ and $0\lt \left|\mathbf{x} - \mathbf{a}\right|\lt \delta$ then $\left| f(\mathbf{x}) - L\right|\lt\epsilon$.

In practice, this definition requires more care than when we working with parametrized functions because limits cannot be computed component-wise as they were previously. This is because of the multidimensional nature of the inputs. We have both $x$ and $y$ (and maybe several others) as independent variables, which makes things a little more involved. Keep this in mind, and we will explore it further later in the lecture.

14.2.2 Visualizations of Limits

Taking the case of a function of two variables, by definition we specify an $\epsilon \gt 0$ that sets the error bound for our function. The corresponding $\delta$ value is the radius of a disk centered at the point in which we want to find the limit.

Here we see $D_\delta$ the disk of radius $\delta$ that forms our neighborhood around $(a,b)$. On the surface, the value of the limit $L$ has to be in the interval $[L-\epsilon,L+\epsilon]$ for each $\epsilon \gt 0$. As $\epsilon$ decreases, so will $\delta$, which will reduce the size of the disk. If we can find a $\delta$ disk that works for every choice of $\epsilon$, then the limit exists.

Paths of Approach

Recall that for limits involving one independent variable, there were two possible paths of approach. If we wanted to determine

$$ \lim_{t\rightarrow t_0} f(t) $$
we had two directions of approach to consider, the limit from the left and the limit from the right . Furthermore, we defined the two-sided limit above to exist if and only if both of these one-sided limits existed and agreed.

In the case of several independent variables, we have a much more sensitive situation. The figure below demonstrates many possible paths to approach a point $(a,b)$.

For a limit involving several variables, every path must result in the same limit .

Therefore, to show that the limit at a point does not exist, it suffices to show that two paths $C_1$ and $C_2$ result in different limits.

14.2.3 Examples of Limits that Do Not Exist

Example 14.2.1

Consider the limit

$$ \lim_{(x,y)\rightarrow (1,0)} \dfrac{xy-y}{(x-1)^2 + y^2} $$

Solution

First let us consider the limit along the $x$-axis. On this line $y=0$, so we have $f(x,0)$ and

$$ \lim_{(x,0)\rightarrow (1,0)} \dfrac{x(0)-(0)}{(x-1)^2 + (0)^2} = \lim_{(x,0)\rightarrow (1,0)} \dfrac{0}{(x-1)^2} = 0 $$
Next, suppose we approach on the line $y = x-1$. Then we have $f(x,x-1)$ and the limit is

$$ \lim_{(x,x-1)\rightarrow (1,0)} \dfrac{x(x-1)-(x-1)}{(x-1)^2 + (x-1)^2} = \lim_{(x,x-1)\rightarrow (1,0)} \dfrac{(x-1)^2}{2(x-1)^2} = \lim_{(x,x-1)\rightarrow (1,0)} \dfrac{1}{2} = \dfrac{1}{2} $$

Since the limits do not match, the limit does not exist. This surface is plotted below.

Example 14.2.2

Consider the limit

$$ \lim_{(x,y)\rightarrow (0,0)} \dfrac{y^2\sin^2 x}{x^4 + y^4} $$

Solution

First consider the limit along the path $x = 0$, then

$$f(x,0) = \frac{y^2\sin^2 (0)}{y^4} = 0$$
for all $x\neq 0$. Hence, the limit is $0$.

Next, consider the limit along the path $y = x$, then

$$f(x,x) = \frac{x^2\sin^2 x}{2x^4} = \frac{1}{2}\left(\frac{\sin x}{x}\right)^2 $$
for all $x\neq y$. We know that $\displaystyle\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$, so the limit along this path is $\frac{1}{2}$. The limits do not agree, so the limit does not exist.

14.2.4 Examples of Limits Which Exist

Exercise 14.2.1

Evaluate the limit


$$ \lim_{(x,y)\rightarrow (0,0)} \dfrac{x^3 - y^3}{x^2 + xy + y^2} $$

Check Your Work

We begin by noticing that the expression in the numerator may be factored.

$$ x^3 - y^3 = (x-y)(x^2 + xy + y^2) $$
Conveniently, one of these factors is precisely the denominator in our limit expression. This expression is zero only when $(x,y) = (0,0)$, so we may use division to eliminate the shared factor. Hence the limit is given by

$$ \begin{align*} \lim_{(x,y)\rightarrow (0,0)} \dfrac{x^3 - y^3}{x^2 + xy + y^2} &= \lim_{(x,y)\rightarrow (0,0)} \dfrac{(x - y)(x^2 + xy + y^2)}{x^2 + xy + y^2} \\ \\ &= \lim_{(x,y)\rightarrow (0,0)} x - y \\ \\ &= 0 \end{align*} $$

Exercise 14.2.2

Evaluate the limit

$$ \lim_{(x,y)\rightarrow (0,0)} \dfrac{x^2 + y^2}{\sqrt{x^2 + y^2+1} - 1} $$

Check Your Work

The denominator for this expression features an inconvenient radical, but we can eliminate it by multiplying by the conjugate of the expression. Doing this will give us an easier to deal with expression for the limit.

$$ \begin{align*} \dfrac{x^2 + y^2}{\sqrt{x^2 + y^2+1} - 1} &= \dfrac{x^2 + y^2}{\sqrt{x^2 + y^2+1} - 1}\cdot\color{#307FE2}{\dfrac{\sqrt{x^2 + y^2+1} + 1}{\sqrt{x^2 + y^2+1} + 1}} \\ \\ &= \dfrac{(x^2 + y^2)(\sqrt{x^2 + y^2+1} + 1)}{x^2 + y^2+1 - 1} \\ \\ &= \dfrac{(x^2 + y^2)(\sqrt{x^2 + y^2+1} + 1)}{x^2 + y^2} \\ \\ &= \sqrt{x^2 + y^2+1} + 1 \end{align*}$$
This expression is very easy to compute the limit of due to continuity . Hence,

$$ \lim_{(x,y)\rightarrow (0,0)} \dfrac{x^2 + y^2}{\sqrt{x^2 + y^2+1} - 1} = \lim_{(x,y)\rightarrow (0,0)} \sqrt{x^2 + y^2+1} + 1 = \sqrt{1} + 1 = 2 $$

We haven't officially discussed continuity in this section, but it works the way you are hoping it does.

Exercise 14.2.3

Evaluate the limit

$$ \lim_{(x,y)\rightarrow (0,0)} \dfrac{xy^4}{x^4 + y^4} $$

Check Your Work

We expect the value of this limit to be zero. However, there are not any algebraic features of this expression that can be easily taken advantage of like the previous exercises. For this, we will be using the Squeeze Theorem .
The most useful fact for us in this problem is that the expression

$$ 0 \le \dfrac{y^4}{x^4 + y^4} \le 1 $$
From here, it must be the case that after multiplication by $|x|$ we have

$$ 0 \le \dfrac{|x|y^4}{x^4 + y^4} \le |x| $$
which is how we will apply the Squeeze Theorem. As $(x,y)\rightarrow (0,0)$, $|x|\rightarrow 0$, so

$$ \begin{align*} 0 \le \lim_{(x,y)\rightarrow (0,0)} \dfrac{|x|y^4}{x^4 + y^4} &\le \lim_{(x,y)\rightarrow (0,0)} |x| \\ \\ 0 \le \lim_{(x,y)\rightarrow (0,0)} \dfrac{|x|y^4}{x^4 + y^4} &\le 0 \end{align*} $$
The limit is zero as this result holds for both positive and negative $x$.

14.2.5 Continuity

As in previous courses, we define continuity as the property of a function being equal to its limit at a point.

Definition

Continuity for a Multivariate Function
A function $f$ of two (or more) variables is continuous at a point $\mathbf{a}$ if

$$ \lim_{\mathbf{x}\rightarrow\mathbf{a}} f(\mathbf{x})\rightarrow f(\mathbf{a}) $$
Furthermore, we say that $f$ is continuous on a domain $D$ if $f$ is continuous at each point $\mathbf{a}\in D$.

Our understanding of continuity extends fairly naturally from our knowledge of one-dimensional functions. Discontinuities in a graph arise from there being a hole or jump in a surface. We have seen several instances of these jumps in examples above where the limit at a point does not exist. In most cases, however, as long as there are not the normal problems of trying to take the square root of a negative number or having zero in a denominator the expressions we work with will be continuous.

Of particular note are polynomials . The expression $x^2 - 2xy + y^2 -5y + 2$ is a polynomial of two variables $x$ and $y$. It, similarly to polynomials of one variable, is continuous in all of its domain $\mathbb{R}^2$. This, as you know from experience, will save us a lot of time computing limits.

Example 14.2.3

Evaluate

$$ \lim_{(x,y)\rightarrow (3,2)} \left(x^2 y^3 - 4y^2\right) $$

Solution

This is a polynomial, so

$$ \lim_{(x,y)\rightarrow (3,2)} \left(x^2 y^3 - 4y^2\right) = \left(3\right)^2\left(2\right)^3 - 4\left(2\right)^2 = 72 - 16 = 56 $$

Example 14.2.4

Determine where the function

$$ f(x,y) = \left\{\begin{matrix} \dfrac{xy^4}{x^4 + y^4} & \quad\text{if }(x,y)\neq (0,0) \\ 0 & \quad\text{if }(x,y) = (0,0) \end{matrix}\right. $$
is continuous.

Solution

This is a rational function for all points except for $(0,0)$, and so it is continuous at all those points. Also, we know that the limit as $f$ approaches the origin is $0$ from Exercise 14.2.3. Therefore, this function is continuous on $\mathbb{R}^2$.

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