Continuity with two variables.

Question

If I'm given a function, say the function $\text f(x,y)=\begin{cases} x\cdot \sin(\frac{1}{y}) + y\cdot \cos(\frac{1}{x}) & \text{if } x\neq 0 \space \text{ and }\space y\neq0 \\ 0 & \text{if } x= 0 \space \text{ or }\space y=0 \\ \end{cases}$

and want to check for continuity, I would at first look at the upper part of which I could say that it is a composition of continuous functions and therefore continuous. For the part defined for one variable being equal to zero, I would look at the limit of my function as $x$ or $y$ approach zero and try to check if that limit is equal to zero, but that's where it gets tricky for me. If, for example I let $y$ go to zero, I'd be left with a part containing an $x$ that isn't defined. Does that mean my limit doesn't exist and my function is not continuous? Thanks for any tips or help.

Answer 1

The conditions $x=0$ and $y=0$ represent the $y-$axis and $x-$axis respectively. We will first consider the limits at different points on the $x$ axis, excluding the point $(0,0)$:

$$\lim_{(x,y)\to(a,0)} \left[x\cdot\sin\left(\frac{1}{y}\right) + y\cdot \cos\left(\frac{1}{x}\right)\right]$$ for $a\neq0$. If we approach the point $(a,0)$ along a line $x-a=ky$ we get the limit

$$\lim_{y\to0}\left[(ky+a)\sin\left(\frac{1}{y}\right) + y\cdot \cos\left(\frac{1}{ky+a}\right)\right] = \text{undefined}$$

Analogously, you can show that the limits at points on the $y-$axis (excluding $(0,0)$ ) do not exist. Thus, we conclude that $f$ is not continuous in these points.

The function is continuous at $(0,0)$, as another answer shows.

Answer 2

Note that, if $x,y\neq0$,\begin{align}\bigl\lvert f(x,y)\bigr\rvert&=\left\lvert x\sin\left(\frac1y\right)+y\cos\left(\frac1x\right)\right\rvert\\&\leqslant\lvert x\rvert+\lvert y\rvert\\&\leqslant2\sqrt{x^2+y^2}.\end{align}Besides, the inequality $\bigl\lvert f(x,y)\bigr\rvert\leqslant2\sqrt{x^2+y^2}$ also holds if $x=0$ or $y=0$. So, given $\varepsilon>0$, take $\delta=\frac\varepsilon2$. Then$$\bigl\lVert(x,y)\bigr\rVert<\delta\iff\sqrt{x^2+y^2}<\delta\implies\bigl\lvert f(x,y)\bigr\rvert<\varepsilon.$$