Question

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.

Answer 1

Given,

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

Thus, partial derivative of the given function exists.

Continuity,

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

Since, limit does not exists.

Therefore, the given function is not continuous.