Section 4.2 - Limits and Continuity

Section Objectives

1. Compute the limit of a multi-variable function.
2. Use the two-path test to show that a limit does not exist.
3. Determine the points of discontinuity of a function of several variables.

Open sets in the plane

Before we can discuss the calculus of functions of several variables, we need to better understand the "structure" (or topology) of multi-dimensional space. Let's focus on two-dimensions. Here are some important words and phrases.

An open disk in $\mathbb{R}^2$ is the interior of a circle, not including the boundary. More specifically, an open disk centered at $(a,b)$ is a set of points satisfying the inequality

for some $r>0$.

An open set in $\mathbb{R}^2$ is a union of open disks. Every point in an open set is an interior point. (The boundary points of a set are not interior points!) We will use the word region to mean an nonempty, open, connected subset of $\mathbb{R}^2$.

A neighborhood of the point $(a,b)$ is any set that entirely contains an open disk centered at $(a,b)$, regardless of disk's radius.

Limits

You began your study of limits in Calculus I by thinking of the limit in an informal way. We will do the same thing here. The following informal defintion is written for two-variable functions, but it generalizes to any number of variables.

Suppose $f(x,y)$ is defined everywhere in a neighborhood of the point $(a,b)$, except possibly at the point $(a,b)$ itself. If $f(x,y)$ gets closer and closer to $L$ as $(x,y)$ gets closer and closer to $(a,b)$, we say

The limits laws that you learned in Calculus I continue to apply with the obvious modifications.

Examples

• Evaluate the limit: $\qquad \displaystyle \lim_{(x,y) \to (1,2)} (x^2+xy-y^2+7)$

• Evaluate the limit: $\qquad \displaystyle \lim_{(x,y) \to (-3,5)} \left( \frac{2x-5y}{x^2+y^2} \right)$

• Evaluate the limit: $\qquad \displaystyle \lim_{(x,y) \to (-3,5)} \left( \frac{xy+3y-2x-6}{x+3} \right)$

• Evaluate the limit: $\qquad \displaystyle \lim_{(x,y) \to (-1,0)} \left[ \frac{2x^2-7y}{(x+1)^2+y^2} \right]$

• Sometimes converting to polar coordinates can be helpful. For example, evaluate the following limit by first converting to polar coordinates: $\quad \displaystyle \lim_{(x,y) \to (0,0)} \frac{\sin(x^2+y^2)}{5x^2+5y^2}$.

Limits at boundary points

In the informal definition of limit given above, we required the function to be defined everywhere around the limit point (except possibly at the limit point). For limits in Calculus I, this was a strong, inflexible requirement. For example, $\lim_{x \to 0} \sqrt{x}$ cannot exist because $\sqrt{x}$ is not defined to the left of $x=0$. For functions of several variables, we may loosen this requirement and allow limits at domain boundary points.

When evaluating a limit at a domain boundary point, we simply take it for granted that the limit point is approached from inside the domain.

For example, the point $(0,1)$ is on the boundary of the domain of the $f(x,y)=\sqrt{1-x^2-y^2}$. To think about the values of $f$ as $(x,y)$ approaches $(0,1)$ clearly requires us to get closer and closer to $(0,1)$ from inside the domain of $f$. In this case, the limit laws apply and the limit is $0$.

• Evaluate the limit: $\qquad \displaystyle \lim_{(x,y) \to (2,2)} \left( \frac{x^2-y^2}{x-y} \right)$

• Evaluate the limit: $\qquad \displaystyle \lim_{(x,y) \to (0,0)} \left( \frac{xy-y^2}{\sqrt{x}-\sqrt{y}} \right)$

Two-path test

In Calculus I, we learned that the "regular" limit was automatically two-sided, and at that point became natural to think about single-sided limits. In Calculus III, the situation far more complicated---there are infinitely many ways to get to a limit point, not just from the left or right. This idea was subtly touched on above when we discussed limits at boundary points. But just as in Calculus I, if a limit is to exist, it should exist and have the same value regardless of how we get to the limit point.

Theorem: If $\displaystyle \lim_{(x,y) \to (a,b)} f(x,y)=L$, then the limit of $f$ is $L$ along any smooth curve that passes through $(a,b)$ and lies in the domain of $f$.

If one could find two different paths to $(a,b)$ that yield two different limits, then from the theorem, the limit cannot possibly exist. We will call this the two-path test.

For example, think about $\displaystyle \lim_{(x,y) \to (0,0)} \frac{2x-y^2}{2x^2+y}$. (Did you notice that $(0,0)$ is a domain boundary point?)

Along the vertical line $x=0$, this limit reduces to $\displaystyle \lim_{y \to 0} \frac{-y^2}{y} = \lim_{y \to 0} (-y) = 0$. On the other hand, along the line $y=x$, the limit reduces to $\displaystyle \lim_{x \to 0} \frac{2x-x^2}{2x^2+x}$, which is equal to $2$. Two different paths give two different limits! The original limit does not exist.

Examples

• Show that the limit does not exist: $\quad \displaystyle \lim_{(x,y) \to (0,0)} \, \frac{2xy}{3x^2+y^2}$.

• Show that the limit does not exist: $\quad \displaystyle \lim_{(x,y) \to (2,1)} \, \frac{(x-2)(y-1)}{(x-2)^2+(y-1)^2}$.

• Find the limit or show that it does not exist: $\quad \displaystyle \lim_{(x,y) \to (0,0)} \, \frac{x^2y}{x+y}$. (This one is tough! Try the path $y=x^3-x$.)

Formal definition of limit

On occasion, we may have to resort to using a more formal definition of limit. Our formal definition is a natural generalization of the $\epsilon$-$\delta$ definition that you may have seen in Calculus I.

Definition: Suppose that $f(x,y)$ is defined in a neighborhood of the point $(a,b)$, except possibly at the point $(a,b)$ itself, or suppose that $(a,b)$ is a boundary point of the domain of $f$. Then

if for any $\epsilon >0$, there exists a corresponding $\delta > 0$ such that for point $(x,y)$ in the domain of $f$,

Examples

• Use the formal definition to show that $\displaystyle \lim_{(x,y) \to (0,0)} \frac{x^2y}{x^2+y^2}=0$. (Can you think of another way to determine this limit?)

Continuity

Now that we understand limits, we can define continuity in terms of limits just as we did in Calculus I. Specifically, we say $f(x,y)$ is continuous at $(a,b)$ if $\displaystyle \lim_{(x,y) \to (a,b)} f(x,y) = f(a,b)$. This definition tells us that a function is continuous at any point where we can get the limit by direct substitution. This is analogous to the definition you saw in Calculus I (except that this new definition also includes continuity at boundaries). Therefore, the properties of single-variable continuous functions generalize very nicely to multi-variable functions.

Examples

• Discuss the continuity of $\displaystyle f(x,y)=\frac{3x+2y}{x+y+1}$.

• Referring to an example above, argue that the only discontinuity of $\displaystyle f(x,y)= \frac{\sin(x^2+y^2)}{5x^2+5y^2}$ is removable.