# [SOLVED]Limit x and y

#### dwsmith

##### Well-known member
$f(x,y) = \begin{cases}(x + y)\sin\frac{1}{x}\sin\frac{1}{y}, & \text{if } x\neq 0\text{ and } y\neq 0\\ 0, & \text{if } x = 0\text{ or } y = 0\end{cases}$

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

#### Sudharaka

##### Well-known member
MHB Math Helper
$f(x,y) = \begin{cases}(x + y)\sin\frac{1}{x}\sin\frac{1}{y}, & \text{if } x\neq 0\text{ and } y\neq 0\\ 0, & \text{if } x = 0\text{ or } y = 0\end{cases}$

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'Hopitals 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.