We will need to consider derivatives of functions of more than one variable. To do this, we first check how the familiar concepts of limits and continuity extend to functions of more than one variable. This material is covered in Section 14.2 of Stewart.

Let $f:D\to\R$ be a function with domain $D$ an open subset of $\R$. For $a\in D$ we say that the limit $\displaystyle{\lim_{x \to a} f(x)}$ exists if and only if, (i) the limit from the left exists, (ii) the limit from the right exists, and (iii) these two limits coincide, i.e., \[ \lim_{x \to a^-} f(x)=\lim_{x \to a^+} f(x). \]

Let $f:D\to\R$ be a function with domain $D$ an open subset of $\R$. For $a\in D$ we say that the limit $\displaystyle{\lim_{x \to a} f(x)}$ exists if and only if, (i) the limit from the left exists, (ii) the limit from the right exists, and (iii) these two limits coincide, i.e., \[ \lim_{x \to a^-} f(x)=\lim_{x \to a^+} f(x). \]

Furthermore, if the limit exists and is equal the actual value of $f$ at $a,$ i.e., if \[ \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a), \] we say that $f$ is continuous at $x=a.$

If $f$ is continuous on all of $D$ we say that $f$ is a continuous function on $D$.

We can also consider the limit for points on the boundary of the domain $D$ of a function. For example, if $f:(0,2)\to\R$ is defined by $f(x)=1/x,$ then \[ \lim_{x\to 2^{-}}f(x)=\frac{1}{2} \]

but $ \ds \lim_{x\to 0^{+}} f(x) $ does not exist.

Another instructive example is $f:\R\backslash\{0\}\to \R$ given by $f(x)=x^2.$ The domain is now the punctured real line, i.e., $D=(-\infty,0)\cup(0,\infty),$ but \[ \lim_{x\to 0^-} f(x)=\lim_{x\to 0^+} f(x)=0. \]

In this situation we also say that $\ds\lim_{x\to 0} f(x)$ exists and in fact one can fix the hole by defining $f(0)=0,$ to extend $f$ to a continuous function on all of $\R.$

Important remark: Never, ever compute $\ds\lim_{x\to a}f(x)$ by blindly substituting $\,x=a\,$ in $\,f.$ For example, if \[ f(x)=\begin{cases} x^2 & \text{for $x\neq 3$} \\ \pi & \text{for $x=3$}, \end{cases} \]

then $\ds \lim_{x\to 3} f(x)$ exists and is given by $9$ which is not equal to $\pi$: the function $f$ is not continuous at $x=3$.

As a second example, if $f:D\to \R$ with $D=\R\backslash\{1\}$ is given by \[ f(x)=\frac{x^2-1}{x-1}, \] then $\ds \lim_{x\to 1}f(x)=2$.

Those (and there will be some) who write "This limit gives $0/0$ which does not exist" should hang their heads in shame.

When $f$ is a function of more than one variable, the situation is more interesting. There are more than two ways to approach a given point of interest. Consider the function \[ f(x,y)=\frac{x^2}{x^2+y^2} \] with domain given by $\R^2\backslash\{(0,0)\}$.

To see the graph of $f$ in Matlab, type

Next we consider the limit as $(x,y)\to(0,0)$.

(i) Approaching the origin along $y=0$:

(ii) Approaching the origin along $x=0$:

In general, for the limit $\displaystyle{ \lim_{(x,y) \to (a,b)} f(x,y)}$ to exist, it is necessary that every path in $D$ approaching $(a,b)$ (the point $(a,b)$ itself may or may not be in $D$) gives the same limiting value. This gives us the following method for finding if a limit does not exist.

Important remark: The above notation is somewhat deficient and perhaps one should write \[ \displaystyle{\lim_{(x,y) \rightarrow_D (a,b)}f(x,y)} \] to indicate that only paths in $D$ terminating in $(a,b)$ (which itself may or may not be in $D$) are considered. For example, if $f(x,y)=x^2+y^2$ with $D=\{(x,y):~x^2+y^2\lt 1\}$ then $\lim_{(x,y)\to (1,0)} f(x,y)$ exists and is $1$.

Important remark: The above notation is somewhat deficient and perhaps one should write \[ \displaystyle{\lim_{(x,y) \rightarrow_D (a,b)}f(x,y)} \] to indicate that only paths in $D$ terminating in $(a,b)$ (which itself may or may not be in $D$) are considered. For example, if $f(x,y)=x^2+y^2$ with $D=\{(x,y):~x^2+y^2\lt 1\}$ then $\lim_{(x,y)\to (1,0)} f(x,y)$ exists and is $1$.

However, if \[ f(x,y)=\begin{cases} x^2+y^2 & \text{for $D=\{(x,y):~x^2+y^2\lt1\}$} \\ 0 & \text{for $D=\{(x,y):~x^2+y^2\gt1\}$} \end{cases} \] then $\lim_{(x,y)\to (1,0)} f(x,y)$ does not exist.

Let $D=\R\backslash\{(0,0)\}$ and $f:D\to \R$ be given by $\displaystyle f(x,y)=\frac{x^2-y^2}{x^2+y^2}. $ Show that $\lim_{(x,y)\to (0,0)} f(x,y)$ does not exist.

With the same $D$ as above but now $\displaystyle f(x,y)=\frac{xy}{x^2+y^2}, $ show that $\lim_{(x,y)\to (0,0)} f(x,y)$ does not exist.

Important remark: There are infinitely many paths terminating in a given point, say $(a,b),$ in $\R^2,$ raising the question if one can ever prove that $\lim_{(x,y)\to (a,b)}f(x,y)$ does exist. The good news is that there are methods that can deal with infinitely many paths simultaneously. The good news is that there are methods that can deal with infinitely many paths simultaneously. The bad news is that these methods (typically $\epsilon$-$\delta$ proofs) are not part of this course.

Important remark: There are infinitely many paths terminating in a given point, say $(a,b),$ in $\R^2,$ raising the question if one can ever prove that $\lim_{(x,y)\to (a,b)}f(x,y)$ does exist. The good news is that there are methods that can deal with infinitely many paths simultaneously. The good news is that there are methods that can deal with infinitely many paths simultaneously. The bad news is that these methods (typically $\epsilon$-$\delta$ proofs) are not part of this course.

See Stewart Sec 14.2, Example 4 for a rigorous $\epsilon$-$\delta$ proof that

where

and $D=\R^2\backslash\{(0,0)\}$.

Example: Give a non-rigorous proof that the above limit is indeed correct by writing $x=r\cos\theta$ and $y=r\sin\theta$.

Example: Give a non-rigorous proof that the above limit is indeed correct by writing $x=r\cos\theta$ and $y=r\sin\theta$.

Definition. Let $f:D\to\R$ be a function with domain $D,$ an open subset of $\R^2.$ Let $(a,b)\in D.$ Then $f(x,y)$ is continuous at $(a,b)$ if \[ \lim_{(x,y) \to (a,b)}f(x,y)=f(a,b), \]

i.e., the limit $(x,y)\to (a,b)$ of $f(x,y)$ exists and is equal to $f(a,b).$

If a function is continuous on all of $D$ we say simply that it is continuous on $D.$ Most of the functions we will consider are continuous. For example, polynomials in $x$ and $y$ are continuous on $\R^2.$ As a rule of thumb, if a function with domain $D$ is defined by a single expression it will be continuous on $D.$

Example: Returning to the first example, where $\displaystyle f(x,y) =\frac{x^2 - y^2}{x^2 + y^2}$ and $D=\R^2\backslash\{(0,0)\},$ is $f(x,y)$ a continuous function?

Example: If we edit the above example by instead defining $f$ on all of $\R^2$ by taking $f(0,0)=0,$ then is $f$ a continuous function?

• You should be able to show when a limit does not exist.
• You should understand continuity of multivariate functions.