Question #107857
check whether the function f (x,y)={4x^2y/4x^4+y^2 , (x,y)not=to (0,0) 0 (x,y)= to (0,0). } Is continuous at (0,0) ? Justify your answer.
1
Expert's answer
2020-04-06T13:00:36-0400
f(x,y)={4x2y4x4+y2if (x,y)(0,0)0if (x,y)=(0,0)f(x,y) = \begin{cases} \dfrac{4x^2y}{4x^4+y^2} &\text{if } (x,y)\not=(0,0) \\ 0 &\text{if } (x,y)=(0,0) \end{cases}

lim(x,y)(0,0)f(x,y)=[y=x2]=lim(x,y)(0,0)4x2(x2)4x4+(x2)2=\lim\limits_{(x,y)\to (0,0)}f(x,y)=[y=x^2]=\lim\limits_{(x,y)\to (0,0)}\dfrac{4x^2(x^2)}{4x^4+(x^2)^2}=

=450=f(0,0)={4\over 5}\not=0=f(0,0)

Therefore the function f(x,y)f(x,y) is not continuous at (0,0).(0,0).



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