Continuity of Functions of Several Variables Examples 1

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

We will now look at some more difficult questions regarding the continuity of functions of several variables.

Example 1

Consider the two variable real-valued function $f(x, y) = \frac{x^2 + y^2 - x^3y^3}{x^2 + y^2}$ for all $(x, y) \in \mathbb{R}^2 \setminus \{ (0, 0) \}$. Determine whether $f$ can be redefined as $\hat{f}$ at the origin such that $\hat{f}$ is continuous for all $(x, y) \in \mathbb{R}^2$.

We first note that $f$ is continuous for all $(x, y) \in \mathbb{R}^2$ such that $x^2 + y^2 \neq 0$ since $f$ is a rational function. In particular, we see that $x^2 + y^2 = 0$ if and only if $(x, y) = (0, 0)$. Hence, the only point of discontinuity occurs at $(0, 0)$. If we can show that $\lim_{(x, y) \to (0, 0)} f(x, y) = L$ for some $L \in \mathbb{R}$ and then define $\hat{f}(0,0) = L$ then we indeed can redefined $f$ at the origin as $\hat{f}$ such that $\hat{f}$ is continuous on all of $\mathbb{R}^2$.

Note that:

(1)

\begin{align} \quad \lim_{(x, y) \to (0,0)} \frac{x^2 + y^2 - x^3y^3}{x^2 + y^2} = \lim_{(x, y) \to (0,0)} \left ( 1 - \frac{x^3y^3}{x^2 + y^2} \right ) \end{align}

Now notice that $\lim_{(x, y) \to (0,0)} \frac{x^3y^3}{x^2 + y^2} = 0$ since:

(2)

\begin{align} \quad 0 ≤ \biggr \rvert \frac{x^3y^3}{x^2 + y^2} \biggr \rvert = \frac{x^2}{x^2 + y^2} \mid x \mid \mid y^3 \mid ≤ \mid x \mid \mid y^3 \mid \end{align}

We have that $\lim_{(x, y) \to (0, 0)} 0 = 0$ and $\lim_{(x, y) \to (0, 0)} \mid x \mid \mid y^3 \mid = 0$, and so by the Squeeze Theorem for multivariable limits we have that $\lim_{(x, y) \to (0,0)} \frac{x^3y^3}{x^2 + y^2} = 0$. Hence:

(3)

\begin{align} \quad \lim_{(x, y) \to (0,0)} \frac{x^2 + y^2 - x^3y^3}{x^2 + y^2} = 1 - 0 = 1 \end{align}

So we will redefine $f$ as follows:

(4)

\begin{align} \quad \hat{f}(x, y) = \left\{\begin{matrix} f(x,y) & \mathrm{if} \: (x, y) \neq (0, 0) \\ 1 & \mathrm{if} \: (x, y) = (0, 0) \end{matrix}\right. \end{align}

Then $\hat{f}$ is continuous for all $(x, y) \in \mathbb{R}^2$. The graph below is of $\hat{f}$.

Example 2

Let $a$, $b$, and $c$ be nonnegative integers. Prove that $\lim_{(x,y) \to (0,0)} \frac{x^ay^b}{(x^2 + y^2)^c}$ exists if $a + b > 2c$.

To show this, note that $\mid x \mid ≤ \sqrt{x^2 + y^2}$ and $\mid y \mid ≤ \sqrt{x^2 + y^2}$. Now applying the Squeeze Theorem for multivariable functions we have that:

(5)

\begin{align} \quad 0 ≤ \biggr \rvert \frac{x^a y^b}{(x^2 + y^2)^c} = \frac{\mid x \mid^a \mid y \mid^b}{(x^2 + y^2)^c} ≤ \frac{\sqrt{x^2 + y^2}^a \sqrt{x^2 + y^2}^b}{(x^2 + y^2)^c} = \frac{(x^2 + y^2)^{\frac{a + b}{2}}}{(x^2 + y^2)^c} = (x^2 + y^2)^{\frac{a + b}{2} - c} \end{align}

Now $\lim_{(x, y) \to (0, 0)} 0 = 0$ and $\lim_{(x, y) \to (0, 0)} (x^2 + y^2)^{\frac{a + b}{2} - c}$ if $\frac{a+b}{2} - c > 0$, that is $a + b > 2c$. So if $a + b > 2c$, we can invoke the Squeeze Theorem and get that:

(6)

\begin{align} \quad \lim_{(x,y) \to (0,0)} \frac{x^ay^b}{(x^2 + y^2)^c} = 0 \end{align}