Question #112039
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)
1
Expert's answer
2020-04-27T17:43:45-0400

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

If the given function is

f(x)={xy2x+2yif(x,y)(0,0)0if(x,y)=(0,0)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 xn=(1n,1n)x_n=(\frac{1}{n},\frac{-1}{n}) in R2\R^2 .

Then limnxn=(0,0).\text{lim}_{n\to \infin}x_n=(0,0). But f(xn)=(1n)(1n)2n2nf(x_n)=\frac{( \frac{1}{n})( \frac{-1}{n} )}{\frac{2}{n}-\frac{2}{n}}

    f(xn)=10\implies f(x_n)=\frac{-1}{0} .

which does not exist.

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


If the function is f(x)={xyx2+y2if(x,y)(0,0)0if(x,y)=(0,0)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 xn=(1n,1n) in R2.x_n=(\frac{1}{n},\frac{1}{n}) \ \text{in} \ \R^2.

Then limxxn=(0,0).\text{lim}_{x\to \infin} x_n=(0,0). But f(xn)=1n22n2=12f(x_n)=\frac{ \frac{1}{n^2}}{\frac{2}{n^2}}=\frac{1}{2} ,which is a constant function and converge to 12\frac{1}{2} .

f(xn)f(0)\therefore f(x_n)\nrightarrow f(0) as nn\rightarrow \infin .

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




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Comments

Assignment Expert
09.05.20, 19:53

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Sheela John
09.05.20, 10:21

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