In our last lecture, we learned to visualize functions of several variables as surfaces. Now, we ask a familiar question from Calculus I: What happens to the output of a function as its input gets closer and closer to a particular point? This is the question of limits.
However, the transition from one dimension to two introduces a profound new complexity. In Calc I, we could approach a point $x=a$ from only two directions: the left or the right. In the $xy$-plane, we can approach a point $(a, b)$ from an infinite number of directions—along straight lines, parabolas, or any other curve. For a limit to exist, the function's output must approach the same value regardless of the path we take. This single idea is the key to everything we will study today.
Let's formalize our intuitive notion of a limit for a function of two variables.
We write $$\lim_{(x, y) \to (a, b)} f(x, y) = L$$ and we say "the limit of $f(x, y)$ as $(x, y)$ approaches $(a, b)$ is $L$" if we can make the values of $f(x, y)$ arbitrarily close to $L$ by taking the point $(x, y)$ to be sufficiently close to the point $(a, b)$, but not equal to $(a, b)$.
The crucial phrase here is "sufficiently close." This implies that no matter how you approach $(a, b)$, the $z$-value of the surface must be honing in on a single height, $L$. If different paths of approach lead to different heights, the limit does not exist.
The precise definition states: $\lim_{(x, y) \to (a, b)} f(x, y) = L$ if for every number $\epsilon > 0$ there is a corresponding number $\delta > 0$ such that if $(x,y)$ is in the domain of $f$ and $0 < \sqrt{(x-a)^2 + (y-b)^2} < \delta$, then $|f(x,y) - L| < \epsilon$.
Geometrically, this means that for any small horizontal strip around the line $z=L$ (of width $2\epsilon$), we can find a small circular disk in the $xy$-plane centered at $(a,b)$ (of radius $\delta$) such that all points on the surface above this disk lie within that horizontal strip. We won't be using this definition to prove limits, but it's the rigorous foundation of the concept.
If we find two distinct paths of approach to $(a,b)$ along which $f(x,y)$ has different limits, what can we conclude about $\lim_{(x, y) \to (a, b)} f(x, y)$?
The most common technique for showing a limit does not exist is the "two-path test." We compute the limit along two different paths. If the results are different, the limit does not exist (DNE).
Show that $\displaystyle \lim_{(x, y) \to (0, 0)} \frac{x^2 - y^2}{x^2 + y^2}$ does not exist.
Notice the chaotic behavior of the surface near the origin.
Let's approach the origin $(0,0)$ along two different lines.
Path 1: The x-axis ($y=0$). Along the x-axis, we have $y=0$. Substituting this into the function, we get: $$ \lim_{(x, 0) \to (0, 0)} \frac{x^2 - 0^2}{x^2 + 0^2} = \lim_{x \to 0} \frac{x^2}{x^2} = \lim_{x \to 0} 1 = 1 $$
Path 2: The y-axis ($x=0$). Along the y-axis, we have $x=0$. Substituting this gives: $$ \lim_{(0, y) \to (0, 0)} \frac{0^2 - y^2}{0^2 + y^2} = \lim_{y \to 0} \frac{-y^2}{y^2} = \lim_{y \to 0} -1 = -1 $$
Since we found two different paths that yield two different limits (1 and -1), we conclude that the limit does not exist.
On the other hand, if a limit does exist, we can't prove it by checking paths. We must use a more rigorous method, like the Squeeze Theorem.
Find $\displaystyle \lim_{(x, y) \to (0, 0)} \frac{3x^2y}{x^2 + y^2}$.
Click on the other formulas to see how the Squeeze Theorem would work.
Testing paths like $y=0$ and $x=0$ both give a limit of 0. This suggests the limit is 0, but doesn't prove it. Let's use an inequality. Note that $x^2 \le x^2 + y^2$, which means $\frac{x^2}{x^2+y^2} \le 1$ for $(x,y) \neq (0,0)$.
Now consider the absolute value of our function:
$$ \left| \frac{3x^2y}{x^2 + y^2} \right| = \frac{3x^2|y|}{x^2+y^2} = 3|y| \left( \frac{x^2}{x^2+y^2} \right) \le 3|y| $$So, we have established that $0 \le \left| \frac{3x^2y}{x^2 + y^2} \right| \le 3|y|$. As $(x,y) \to (0,0)$, we know that $|y| \to 0$, so $3|y| \to 0$. By the Squeeze Theorem, since our function's absolute value is trapped between 0 and something approaching 0, its limit must also be 0.
Therefore, $\displaystyle \lim_{(x, y) \to (0, 0)} \frac{3x^2y}{x^2 + y^2} = 0$.
Investigate $\displaystyle \lim_{(x, y) \to (0, 0)} \frac{2x^2y}{x^4 + y^2}$.
Click on the last formula and set t to -10 < t < 10.
First, let's check the standard axes.
Path 1: The x-axis ($y=0$). The limit is $\mathbf{0}$.
Path 2: The y-axis ($x=0$). The limit is also $\mathbf{0}$.
The limits along both axes agree! However, this is not sufficient proof. Let's try a non-linear path suggested by the powers in the denominator: the parabola $y=x^2$.
Path 3: The parabola $y=x^2$. $$ \lim_{(x, x^2) \to (0, 0)} \frac{2x^2(x^2)}{x^4 + (x^2)^2} = \lim_{x \to 0} \frac{2x^4}{x^4 + x^4} = \lim_{x \to 0} \frac{2x^4}{2x^4} = \lim_{x \to 0} 1 = \mathbf{1} $$
Since we found paths that yield different limits (0 and 1), the limit does not exist. This example powerfully illustrates that you must consider all paths, not just the easy ones.
Fortunately, we don't always have to resort to path-testing or the Squeeze Theorem. Many of the familiar limit laws from Calculus I have direct analogues for functions of two variables.
| Law | Description |
|---|---|
| Sum/Difference | $\lim [f(x,y) \pm g(x,y)] = \lim f(x,y) \pm \lim g(x,y)$ |
| Constant Multiple | $\lim [c \cdot f(x,y)] = c \cdot \lim f(x,y)$ |
| Product | $\lim [f(x,y) \cdot g(x,y)] = \lim f(x,y) \cdot \lim g(x,y)$ |
| Quotient | $\lim \frac{f(x,y)}{g(x,y)} = \frac{\lim f(x,y)}{\lim g(x,y)}$ (if $\lim g(x,y) \neq 0$) |
| Power | $\lim [f(x,y)]^n = [\lim f(x,y)]^n$ |
A powerful consequence of these laws is that for certain well-behaved functions, we can find the limit by simply plugging in the values.
If $f$ is a polynomial or a rational function and $(a,b)$ is in the domain of $f$, then $$ \lim_{(x,y) \to (a,b)} f(x,y) = f(a,b) $$ This property also holds for many other familiar functions, like trigonometric, exponential, and logarithmic functions, as long as $(a,b)$ is in their domain.
Evaluate $\displaystyle \lim_{(x,y) \to (1, 2)} (x^2y + 3y^2 - 5)$.
The idea of continuity also extends naturally from Calculus I. Intuitively, a function is continuous if its graph is an unbroken surface with no holes or jumps. The formal definition relies on the concept of a limit.
A function $f$ of two variables is called continuous at $(a, b)$ if $$ \lim_{(x, y) \to (a, b)} f(x, y) = f(a, b) $$ This single equation implies three conditions:
Just as with limits, polynomials and rational functions are continuous on their domains. This means we can often determine where a function is continuous by simply finding its domain.
Determine the set of points at which the function $f(x,y) = \frac{\ln(x^2+y^2-4)}{x-3}$ is continuous.
This is a combination of other continuous functions. It will be continuous everywhere on its domain. We need to find the domain by looking for restrictions.
The function is continuous on the set $D = \{(x,y) | x^2+y^2 > 4 \text{ and } x \neq 3\}$. This is the entire plane except for the disk of radius 2 and the vertical line $x=3$.
Consider the function $$ g(x, y) = \begin{cases} \frac{x^2 - y^2}{x^2 + y^2} & \text{if } (x, y) \neq (0, 0) \\ 0 & \text{if } (x, y) = (0, 0) \end{cases} $$ Is $g$ continuous at $(0,0)$?
Let's check the three conditions for continuity at $(0,0)$.
Since the second condition fails, the function is not continuous at $(0,0)$. There is a "tear" or discontinuity in the surface at the origin.
All of these ideas—limits, limit laws, continuity—extend to functions of three or more variables. For a function $f(x, y, z)$, the limit is defined as: $$ \lim_{(x, y, z) \to (a, b, c)} f(x, y, z) = L $$ This means the value of $f(x,y,z)$ approaches $L$ as the point $(x,y,z)$ approaches $(a,b,c)$ along any path in 3D space. Direct substitution works for continuous functions (like polynomials and rational functions) in the same way.
Find $\displaystyle \lim_{(x,y,z) \to (2, 0, -1)} e^{-xy} \sin(\pi z/2)$.
Show that $\displaystyle \lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2 + y^4}$ does not exist.
Where is the function $f(x, y) = \arctan(y/x)$ continuous?
Today we extended the fundamental calculus concepts of limits and continuity to functions of several variables. While the definitions look similar, the move to higher dimensions brings a crucial new challenge: the path of approach matters immensely.
These concepts form the bedrock upon which we will build the differential calculus of multivariable functions in the coming lectures.