# Continuity of Functions of Several Variables Examples 2

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 $f(x, y) = \frac{\sin x \sin^3 y}{1 - \cos (x^2 + y^2)}$. Determine whether $f$ can be redefined as $\hat{f}$ at the origin such that $\hat{f}$ is continuous at the origin.**

If we can show that $\lim_{(x, y) \to (0, 0)} \frac{\sin x \sin^3 y}{1 - \cos (x^2 + y^2)} = L$, then we can define $\hat{f}$ to be $f$ for $(x, y) \in D(f)$, $(x, y) \neq (0, 0)$ and define $\hat{f}$ to be $L$ at $(0, 0)$ so that $\hat{f}$ is continuous at the origin.

Let's try to evaluate $\lim_{(x, y) \to (0, 0)} \frac{\sin x \sin^3 y}{1 - \cos (x^2 + y^2)} = L$. Along the line $y = 0$ we have that:

(1)However, along the line $x = y$ we have that:

(2)We now use the trigonometric identity that $\sin^2 u = \frac{1 - \cos (2u)}{2}$. We then have that $2\sin^2 (y^2) = 1 - \cos (2y^2)$. Plugging this into the limit above and:

(3)Using L'Hospital's rule multiple times and we see that this limit is equal to $\frac{1}{2}$. Therefore $L$ does not exist and $\hat{f}$ cannot be a redefinition of $f$ and be continuous at the origin. The graph of $f$ is given below: