Discontinuities in Functions of Several Variables
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# Discontinuities in Functions of Several Variables

Recall from the Continuity of Functions of Several Variables page that a two variable real-valued function $z = f(x, y)$ is continuous at $(a, b) \in D(f)$ if $\lim_{(x,y) \to (a,b)} f(x, y) = f(a, b)$ and a three variable real-valued function $w = f(x, y, z)$ is continuous at $(a, b, c) \in D(f)$ if $\lim_{(x, y, z) \to (a, b, c)} f(x, y, z) = f(a, b, c)$.

If the above do not hold, then we say that $f$ is Discontinuous at that point in the domain of $f$. We will make some important remarks regarding two conditions which do NOT imply continuity.

 Remark 1: Suppose $z = f(x, y)$ is a two variable real-valued function and $(a, b) \in D(f)$. Then the continuity of $g(x) = f(x, b)$ at $x = a$ and the continuity of $h(y) = f(a, y)$ at $y = a$ does NOT imply the continuity of $f(x, y)$ at $(a, b)$.

Remark 1 is important to understand. We note that $g(x) = f(x, b)$ is the curve we obtain by fixing $y = b$ and letting $x$ vary, that is, $g(x)$ is the curve of intersection of the plane $y = b$ and the surface $z = f(x, y)$. Similarly, $h(y) = f(a, y)$ is the curve of intersection of the plane $x = a$ and the surface $z = f(x, y)$.

Even if both of these curves, call them $C_1$ and $C_2$ are continuous at $x = a$ and $y = b$ respectively is NOT enough to imply the continuity of $f$. We will show this through a counterexample. Define $f(x, y)$ as follows:

(1)
\begin{align} f(x,y) = \left\{\begin{matrix} \frac{xy}{x^2 + y^2} & \mathrm{if} \: (x,y) \neq (0, 0)\\ 0 & \mathrm{if} (x, y) = (0,0) \end{matrix}\right. \end{align}

Consider the point of interest $(a, b) = (0, 0)$ and define $g$ and $h$ as follows:

(2)
\begin{align} \quad g(x) = f(x, b) = f(x, 0) = \frac{x(0)}{x^2 + (0)^2} = 0 \end{align}
(3)
\begin{align} \quad h(y) = f(a, y) = f(0, y) = \frac{(0)y}{(0)^2 + y^2} = 0 \end{align}

Clearly $g(x)$ is continuous at $x = 0$ and $h(y)$ is continuous at $y = 0$. However, we will now show that $f(x, y)$ is not continuous at $(0, 0)$ by showing that $f$ does not approach $0$ along some path in the domain of $f$. Notice that if we take the limit as $(x, y) \to (0, 0)$ along the line $y = x$ we get that:

(4)
\begin{align} \lim_{(x, y) \to (0, 0)}_{\mathrm{along} \: y = x} \frac{xy}{x^2 + y^2} = \lim_{(x, x) \to (0, 0)} \frac{x^2}{x^2 + x^2} = \lim_{x \to 0} \frac{x^2}{2x^2} = \frac{1}{2} \neq 0 \end{align}

Therefore $f$ is not continuous at $(0, 0)$.

 Remark 2: Suppose $z = f(x, y)$ is a two variable real-valued function and $(a, b) \in D(f)$. Let $\vec{u} = (v, w)$ be a unit vector and consider the line with parametric equations $L = \left\{\begin{matrix} a + tv\\ b + tw\end{matrix}\right.$ and define $f_{\vec{u}} (t) = f(a + tv, b + tw)$. Note that $L$ is the line that passes through $(a, b)$ and is parallel to $\vec{u}$. Also note that $f_{\vec{u}} (t)$ is the curve generated by taking the values of $f$ at the points on the line $L$, $(x, y) = (a + tv, b + tw)$ in the domain of $f$. Then the continuity of $f_{\vec{u}} (t)$ at $t = 0$ for every unit vector $\vec{u}$ does NOT imply the continuity of $f(x, y)$ at $(a, b)$.

The conditions in Remark 2 can be seen as an extension to Remark 2. Now we're dealing with looking at the curves of intersection between the surface $z = f(x, y)$ and any plane perpendicular to the $xy$-plane (which is equivalent to the curve $f_{\vec{u}} (t)$ generated from taking points $(x, y) \in D(f)$ along an arbitrary line $L$ defined above).

To show that the continuity of $f_{\vec{u}} (t) = f(a + tv, b + tw)$ at $t = 0$ for all unit vectors $\vec{u}$ does not imply the continuity of $f(x, y)$ at $(a, b)$, consider the following counterexample:

(5)
\begin{align} \quad f(x, y) = \left\{\begin{matrix} \frac{x^2 y}{x^4 + y^2} & \mathrm{if} \: (x, y) \neq (0,0)\\ 0 & \mathrm{if} \: (x, y) = (0, 0) \end{matrix}\right. \end{align}

In this case we will have $(a, b) = (0, 0)$. Every straight line $L$ through the origin parallel to some unit vector $\vec{u}$ can be written as $y = mx$ or $x = 0$.

Suppose that we took the limit of the function above as $(x, y) \to (0, 0)$ first along the lines $y = mx$. Then:

(6)
\begin{align} \quad \lim_{(x, y) \to (0, 0)}_{\mathrm{along} \: y = mx} \frac{x^2 y}{x^4 + y^2} = \lim_{(x, kx) \to (0, 0)} \frac{kx^3}{x^4 + k^2x^2} = \lim_{x \to 0} \frac{kx^3}{x^4 + k^2x^2} = \lim_{x \to 0} \frac{kx}{x^2 + k^2} = 0 \end{align}

Also consider the line $x = 0$:

(7)
\begin{align} \quad \lim_{(x, y) \to (0, 0)}_{\mathrm{along} \: x = 0} \frac{x^2 y}{x^4 + y^2} = \lim_{x \to 0} 0 = 0 \end{align}

Therefore along all straight lines $L$ that pass through $(0, 0)$ from the domain, the corresponding curves $f_{\vec{u}}(t)$ are continuous at $t = 0$.

However, the function $f$ is not continuous at $(0, 0)$ as:

(8)
\begin{align} \quad \lim_{(x, y) \to (0, 0)}_{\mathrm{along} \: y = x^2} \frac{x^2y}{x^4 + y^2} = \lim_{(x, x^2) \to (0, 0)} \frac{x^4}{2x^4} = \lim_{x \to 0} \frac{x^4}{2x^4} = \frac{1}{2} \neq 0 \end{align}
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