Answer to Question #112039 in Real Analysis for Sheela John

Question

If f(X,y)={xy/x2+y2, if(X,y)#(0,0)

0 if (X,y)=(0,0)
Check the continuity CF F(X,y) at (0,0)

Answer 1

I have doubts in the question, so I have given two answers.

If the given function is

$f(x)=\begin{cases} \frac{xy}{2x+2y}&\text{if} (x,y)\neq(0,0) \\ 0 & \text{if} (x,y)=(0,0) \end{cases}$

Consider the sequence $x_n=(\frac{1}{n},\frac{-1}{n})$ in $\R^2$ .

Then $\text{lim}_{n\to \infin}x_n=(0,0).$ But $f(x_n)=\frac{( \frac{1}{n})( \frac{-1}{n} )}{\frac{2}{n}-\frac{2}{n}}$

$\implies f(x_n)=\frac{-1}{0}$ .

which does not exist.

Hence $f(x,y)$ is not continuous at (0,0).

If the function is $f(x)=\begin{cases} \frac{xy}{x^2+y^2} & \text{if} (x,y)\neq(0,0) \\ 0 & \text{if} (x,y)=(0,0) \end{cases}$

Consider the sequence $x_n=(\frac{1}{n},\frac{1}{n}) \ \text{in} \ \R^2.$

Then $\text{lim}_{x\to \infin} x_n=(0,0).$ But $f(x_n)=\frac{ \frac{1}{n^2}}{\frac{2}{n^2}}=\frac{1}{2}$ ,which is a constant function and converge to $\frac{1}{2}$ .

$\therefore f(x_n)\nrightarrow f(0)$ as $n\rightarrow \infin$ .

Hence $f(x,y)$ is not continuos at (0,0).