Question

find fx(0,0) and fx (x,y), where (x,y)≠(0,0) for the following function f(x,y)= {xy^3/x^2+y^2, (x,y) ≠(0,0) 0, (x,y)=(0,0) is fx continuous at (0,0)?

Answer 1

The given equation is

$f(x,y)=\begin {cases} \frac{(xy^3)}{x^2} +y^2 , & \text{if} \ (x,y)\neq(0,0) \\ 0 ,& \text{if} \ (x,y)=(0,0) \end{cases}$

Then

$$f_x(0,0)=\lim_{h\to 0} \frac{f(0+h,0)-f(0,0)}{h}$$ $$=\lim_{h\to0} \frac{(0-0)}{h}$$ $$=0$$

$$\text{Now},\ \ f_x(x,y)=\lim_{h\to0} \frac{f(x+h,y)-f(x,y)}{h}$$ $$=\lim_{h\to0} \frac{(\frac{y^3}{x+h}+y^2)-(\frac{y^3}{x}+y^2)}{h}$$

$$=\lim_{h\to0} \frac{x y^3-xy^3-hy^3}{h(x+h)(x)}$$ $$=\lim_{h\to0} \frac{-hy^3}{h(x+h)(x)}$$

$$=\frac{-y^3}{x^2} \ \ \text{when} \ \ (x,y)\neq(0,0)$$

$\text{Consider the sequence } \ (x_n)=(\frac{1}{n^2},\frac{1}{n}) \in \R^2$ .

Then

$$||(x_n)-(0,0)||=||x_n||=\sqrt{(\frac{1}{n^4}+\frac{1}{n^2})}$$

$$=\sqrt{ \frac{n^2+1}{n^4}}=\frac{ \sqrt{1+n^2}}{n^2}\leq\frac{ \sqrt{1+2n+n^2}}{n^2}$$

$$=\frac{1+n}{n^2}\leq\frac{1}{n^2}+\frac{1}{n}\leq \frac{1}{n}+\frac{1}{n}$$ $$=\frac{2}{n}\leq\epsilon \ , \text{whenever} \ n\geq K$$

$\text{Where}, \ K>\frac{2}{\epsilon}$

$\therefore \ (x_n)\rightarrow(0,0)$ but

$$|f_x(x_n)-f_x(0,0)|=|-\frac{n^4}{n^3}-0|=n\rightarrow\infin \ as \ n\rightarrow\infin$$ $$\text{Hence}, \ f_x(x_n)\nrightarrow f_x(0,0)$$

$\text{Therefore} \ , f_x$ is not continuous at (0,0).