Analyzing Discontinuities of $f(x,y)$

[Dustinsfl]

$f(x,y) = \begin{cases}(x + y)\sin\frac{1}{x}\sin\frac{1}{y}, & \text{if } x\neq 0\text{ and } y\neq 0\\
0, & \text{if } x = 0\text{ or } y = 0\end{cases}$ We can re-write $f$ as
$$
f(x,y) = \begin{cases}
\frac{x + y}{xy}\frac{\sin\frac{1}{x}\sin\frac{1}{y}}{\frac{1}{x}\frac{1}{y}}, & \text{if } x\neq 0\text{ and } y\neq 0\\
0, & \text{if } x = 0\text{ or } y = 0\end{cases}
$$

Can I use l'hospital's rule here? Taking the limit gives 0/0. I am looking for points of discontinuity if any exist in $f$.

 

[Sudharaka]

$f(x,y) = \begin{cases}(x + y)\sin\frac{1}{x}\sin\frac{1}{y}, & \text{if } x\neq 0\text{ and } y\neq 0\\
0, & \text{if } x = 0\text{ or } y = 0\end{cases}$ We can re-write $f$ as
$$
f(x,y) = \begin{cases}
\frac{x + y}{xy}\frac{\sin\frac{1}{x}\sin\frac{1}{y}}{\frac{1}{x}\frac{1}{y}}, & \text{if } x\neq 0\text{ and } y\neq 0\\
0, & \text{if } x = 0\text{ or } y = 0\end{cases}
$$

Can I use l'hospital's rule here? Taking the limit gives 0/0. I am looking for points of discontinuity if any exist in $f$.

Hi dwsmith, :)

L'Hopital's rule is for single variable functions and extensions of it to multivariable functions is rare, but you maybe interested by >>this<<. You can solve this problem using the Sandwitch theorem,

When, \(x\neq 0\) and \(y\neq 0\) we have,

\[f(x,y)=(x + y)\sin\frac{1}{x}\sin\frac{1}{y}=x\sin\frac{1}{x} \sin\frac{1}{y} + y\sin\frac{1}{x}\sin\frac{1}{y}\]

Note that,

\[-1\leq\sin\frac{1}{x}\sin\frac{1}{y}\leq 1\]

\[\Rightarrow -x\leq x\sin\frac{1}{x}\sin\frac{1}{y}\leq x\mbox{ and }-y\leq y\sin\frac{1}{x}\sin\frac{1}{y}\leq y\]

So when both \(x,y\rightarrow 0\) we have,

\[\lim_{(x,y)\rightarrow (0,0)}f(x,y)=0=f(0,0)\]

Therefore \(f\) is continuous at the point \((0,0)\). However if only one variable \(x\) or \(y\) tends to zero and the other one doesn't the limit does not exist. Hence the points of discontinuities are,

\[S=\left\{(0,y)\mbox{ where }y\neq 0\right\}\cup\left\{(x,0)\mbox{ where }x\neq 0\right\}\]

Kind Regards,
Sudharaka.