Simple Continuous Piecewise function

Question

(i) Define $f:R^2 →R$ by $$f(x,y)= \frac {xy} {x^2+y^2}$$ when $(x,y) \ne (0,0)$ $$f(x,y)= 0 $$ when $(x,y)=(0,0)$ Prove f is continuous away from $(0, 0)$ but not continuous at $(0,0)$.

(ii) Let $f:\mathbb{R} \to \mathbb R$ be defined by $f(x)=\sin(x)$ when $x \in \mathbb Q$ and $0$ otherwise. At which values of $x \in \mathbb R$ is the function continuous?

For part (i), I just have to show that when $x$ and $y$ both go to $0$, then the function does not approach $0$ but rather infinity? f is continuous away from $(0,0)$, does that mean I have to show f is continuous at all other points? If so, do I simply say $x^2+y^2$ does not equal $0$ at all other points thus continuous?

For part (ii), I am not entirely sure on how to find the points.

Answer 1

A real function $f$ is continuous at $(a,b)\in\Bbb{R}^2$ if $\lim_{(x,y)\to(a,b)}f(x,y)=f(a,b)$. To show that $f$ is continuous away from $(0,0)$ you should show that it is continuous at every point except (perhaps) the origin.

For the second question, again check the definitions. For which $c\in\Bbb{R}$ do you have $\lim_{x\to c}f(x)=f(c)$? Here's a hint: Every real number is the limit of a sequence of rational numbers.

Answer 2

Taking the limit along the line $y=tx$ (note $x\to0\iff y\to 0$ still holds), we see:

$$\lim_{(x,y)\to(0,0)}f(x,y)=\lim_{x\to0}\frac{tx^2}{x^2+t^2x^2}=\lim_{x\to 0}\frac{t}{1+t^2}=\frac{t}{1+t^2}$$

The limit varies depending on the value of $t$, so the limit $\lim_{(x,y)\to(0,0)}f(x,y)$ does not exist.

Hence $f(x,y)$ is discontinuous at $(0,0)$.

Since $xy,x^2+y^2$ are continuous functions, and $x^2+y^2\neq 0$ outside of $(0,0)$, the quotient of them is continuous outside of $(0,0)$