Answer to Question #108192 in Calculus for Garima Ahlawat

Question

Let f(x,y) = {xy/(x^2 +y^2) , (x,y) not equal to (0,0) , 0 , (x,y) =(0,0)}
(I) show that f(x,y) and f(x,0) are each continuous functions of one variable.
(ii) is f continuous at (0,0)? Give reason for your answer.
(iii) is f differentiable at (0,0)? Give reason.

Answer 1

(i) Let $x=x_0=const\not=0$

$f(x_0,y)={x_0y \over x_0^2+y^2}, y\not=0, f(x_0,0)=0$ $\lim\limits_{(x_0,y)\to(x_0,0)}f(x_0,0)=\lim\limits_{y\to0}{x_0y \over x_0^2+y^2}=0=f(x_0,0)$ $f(0,y)={0(y) \over 0^2+y^2}=0, y\not=0, f(0,0)=0$ $\lim\limits_{(0,y)\to(0,0)}f(x,0)=\lim\limits_{y\to0}\big[0\big]=0=f(0,0)$

The function $f(x_0,y)$ is continuous at $y=0.$

Let $y=y_0=const\not=0$

$f(x,y_0)={xy _0\over x^2+y_0^2}, x\not=0, f(0,y_0)=0$

$\lim\limits_{(x,y_0)\to(0,y_0)}f(x,y_0)=\lim\limits_{x\to0}{xy_0 \over x^2+y_0^2}=0=f(0,y_0)$

$f(x,0)={x(0) \over x^2+0^2}=0, x\not=0, f(0,0)=0$

$\lim\limits_{(x,0)\to(0,0)}f(x,0)=\lim\limits_{x\to0}\big[0\big]=0=f(0,0)$

The function $f(x,0)$ is continuous at $x=0.$

The function $f(x, y_0)$ is continuous at $x=0.$

(ii) Let $x=y$

$\lim\limits_{(x,y)\to(0,0)}f(x,y)=\lim\limits_{x\to0}\big[{x^2\ \over x^2+x^2}\big]={1 \over 2}\not=0=f(0,0)$

The function $f$ is not continuous at (0,0)

(iii)

$f_x(x,y)=y\cdot{x^2+y^2-2x^2 \over (x^2+y^2)^2}=y{y^2-x^2 \over (x^2+y^2)^2},\ (x,y)\not=(0,0)$

$f_y(x,y)=x\cdot{x^2+y^2-2y^2\over (x^2+y^2)^2}=x{x^2-y^2\over (x^2+y^2)^2},\ (x,y)\not=(0,0)$

The partial derivatives are defined at (0, 0).

$f_x(0,0)=\lim\limits_{h\to0}{1\over h}(f(0+h,0)-f(0,0))=\lim\limits_{h\to0}(0-0)=0$

$f_y(0,0)=\lim\limits_{h\to0}{1\over h}(f(0,0+h)-f(0,0))=\lim\limits_{h\to0}(0-0)=0$

Therefore

$f_x(0,0)=f_y(0,0)=0$