Limits of Functions of Two Variables
Let $y = f(x)$ be a function of a single variable. Recall that we say $\lim_{x \to c} f(x) = L$ if $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $0 < \mid x - c \mid < \delta$ then $\mid f(x) - L \mid < \epsilon$. We will now extend the concept of a limit to a function of two variables.
Definition: Let $z = f(x, y)$ be a two variable real-valued function. Then the Limit of $f(x, y)$ as $(x, y)$ Approaches $(a,b)$ is $L$ denoted $\lim_{(x, y) \to (a,b)} f(x, y) = L$ if $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $(x, y) \in D(f)$ and $0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta$ then $\mid f(x, y) - L \mid < \epsilon$. |
One important similarity to notice between the limit of a one variable function and the limit of a two variable function is that $\sqrt{(x - a)^2 + (y - b)^2}$ represents the distance between the point $(x, y)$ and $(a, b)$ in $\mathbb{R}^2$. Also, $\mid f(x, y) - L \mid$ represents the distance between the numbers $f(x, y)$ and $L$ in $\mathbb{R}$ as usual. Thus, another way to formulate the definition above is as follows. $\lim_{(x,y) \to (a,b)} f(x,y) = L$ if $\forall \epsilon > 0$ $\exists \delta > 0$ such that if the point $(x, y)$ is contained in the domain of $f$ and is within a distance of less than $\delta$ but not $0$ from $(a, b)$, then the distance between $f(x,y)$ and $L$ is less than $\epsilon$.
In terms of open disks and open intervals, if $D_{\delta}$ is a open disk centered at $(a, b)$ with radius $\delta > 0$, then for all points $(x, y)$ contained in this disk and the domain of $f$, then $f(x,y)$ is contained in the $\epsilon$-neighbourhood of $L$, i.e, if $(x, y) \in D_{\delta} \cap D(f)$ then $f(x,y) \in B_{\epsilon} (L)$.
Geometrically, the limit of a two variable function at $(a, b)$ is $L$ implies that for all $\epsilon > 0$ then there exists a $\delta > 0$ such that the subsurface generated by the direct image of the open disk centered at $(a, b)$ but not necessarily containing $(a,b)$, i.e, $D_{\delta} \setminus \{ (a, b) \}$ is contained between the planes $z = L + \epsilon$ and $z = L -\epsilon$ as shown in the image below.
Note that this is analogous to functions of one variable, that is $\lim_{x \to c} f(x) = L$ if $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $0 < \mid x - c \mid < \delta$ then $L -\epsilon < f(x) < L + \epsilon$, in other words, the portion of the curve generated by the direct image $f((x - c, x + c) \setminus \{ c \})$ lies between the horizontal lines $y = L + \epsilon$ and $y = L - \epsilon$.
The Limits of a Two Variable Function from Different Directions
Recall that for a function of one variable that the $\lim_{x \to c} f(x) = L$ if and only if $\lim_{x \to c^-} f(x) = L = \lim_{x \to c^+} f(x)$, that is, both the lefthand and righthand limits exist and equal one another. When dealing with two variable functions, this can be an issue since we can approach the point $(a, b)$ from more than two directions - in fact, we can approach this point from infinitely many directions as illustrated below.
For that reason, we establish the following definition for the non-existence of a limit.
Definition: If $z = f(x, y)$ is a two variable real-valued function, then we say $\lim_{(x, y) \to (a, b)} f(x)$ Does Not Exist (often abbreviated D.N.E.) if either: a) There exists a path $C$ where if $(x, y) \to (a,b)$ along $C$ we have that $f(x,y) \to \pm \infty$. b) There exists two paths $C_1$ and $C_2$ for which if when $(x, y) \to (a, b)$ along $C_1$ then $f(x, y) \to L_1$ AND as $(x, y) \to (a, b)$ along $C_2$ then $f(x, y) \to L_2$ where $L_1 \neq L_2$. |
Notice that (b) says that if we can find two paths of points $(x, y)$ to the point $(a, b)$ in the domain of $f$ that approach different values, then we can conclude that $\lim_{(x, y) \to (a, b)} f(x, y)$ does not exist. Note that when we say "find two paths", we will often fix one of the variables. For example, if we wanted to find the value $f$ approaches along the $x$-axis, we would fix $y = 0$ and evaluate the limit which would now be in terms of $x$ only. If we wanted to find the value $f$ approaches along the line $y = x^2$, we would substitute $x^2$ in for $y$ and once again evaluate the limit involving the variable $x$ only.
Also note that the opposite finding does not produce the opposite conclusion. If we find two paths of points $(x, y)$ to $(a,b)$ in the domain of $f$ that approach the same values, this does NOT guarantee us that the limit exists. We would in theory have to determine what value each path that approaches $(a, b)$ takes on. This is impossible to do of course as there are infinitely many paths of infinitely many directions and types. In such a case, we can test some paths, and if they lead us to believe a certain value is the limit as $(x, y) \to (a, b)$, then we will have to use the definition above to prove its existence similarly to how we proved limits of functions of one variable.
Example 1
Show that the following limit does not exist: $\lim_{(x, y) \to (0, 0)} \frac{x^2 + 3y^2}{x^2 + y^2}$.
To show that this limit does not exist, we will show that there exists two paths to the point $(0, 0)$ that approach different values.
Consider the path along the $x$-axis, that is $y = 0$. So as $(x, y) \to (0, 0)$ we have that $f(x, y) = \frac{x^2}{x^2} = 1$, and so $f(x, y) \to 1$.
Now consider the path along the $y$-axis, that is $x = 0$. So as $(x, y) \to (0, 0)$ we have that $f(x, y) = \frac{3y^2}{y^2} = 3$, and so $f(x, y) \to 3$.
Therefore, the $\lim_{(x, y) \to (0,0)} \frac{x^2 + 3y^2}{x^2 + y^2}$ does not exist.
The following graph illustrates the two paths we chose above and that the limit does not exist as $(x, y) \to (0, 0)$.
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