Answer to Question #187991 in Calculus for charan

Question #187991

if function f(x,y)={xy/x^2+y^2, (x,y) is not equal to (0,0) 0 , (x,y)=(0,0) is continuous at (0,0).then show that both the partial derivative exists at the origin, but the function is not continuous at the origin.


1
Expert's answer
2021-05-07T10:22:43-0400

Given,

"function \\newline f(x,y)=\\frac{xy}{x^2+y^2}, (x,y) \\neq (0,0) \\newline\nf(x,y)=0, (x,y)=(0,0)\\newline\n\\text{For partial derivative,}\\newline\nf_{x}(0,0)=lim_{ h \\to 0} \\frac{f(h,0)-f(0,0)}{h}\\newline\n=0 \\space (exists)\\newline\nf_{y}(0,0)=lim_{ k \\to 0} \\frac{f(0,k)-f(0,0)}{k}\\newline\n=0 \\space (exists)"

Thus, partial derivative of the given function exists.

Continuity,

"Let y=mx.\\newline\nlim_{ (x,y) \\to (0,0)} f(x,y)\\newline\n=lim_{ (x,y) \\to (0,0)}\\frac{xy}{x^2+y^2}\\newline\nput y=mx,\\newline\nlim_{ (x,mx) \\to (0,0)}\\frac{mx^2}{x^2+m^2x^2}\\newline\n=lim_{ x\\to 0}\\frac{m}{1+m^2}\\newline\n=\\frac{m}{1+m^2} \\space ( not exists)"

Since, limit does not exists.

Therefore, the given function is not continuous.


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