calculating partial derivatives at $(0,0)$

Question

Let $f:\mathbb R^2 \to \mathbb R$ given by :=

$$f(x,y) = \begin{cases} 0 & \text{, if xy=0 } \\ 1 & \text{, if xy $\neq$ 0} \end{cases}$$

I've to show that $\partial_1 f(0,0)=0=\partial_2 f(0,0)$.

Also show that $f$ is not continuous at $0$..

I don't know how to calculate partial derivatives in this case.please if anyone can explain it to me...

You can show discontinuity by looking at some limit of $f(x,y)$ as $(x,y)$ approaches $(0,0)$. — Bruce Zheng, Oct 16 '14 at 05:13

score 2 · Accepted Answer · answered Oct 16 '14 at 05:38

$\partial_1 f(0,0) = \displaystyle \lim_{h \to 0} \dfrac{f(h,0) - f(0,0)}{h} = \lim_{h \to 0} \dfrac{0 - 0}{h} = 0$.

$\partial_2 f(0,0) = \displaystyle \lim_{k \to 0} \frac{f(0,k) - f(0,0)}{k} = \lim_{k \to 0} \dfrac{0 - 0}{k} = 0$.

For $(x_n,y_n) = (\frac{1}{n}, \frac{1}{n})$, we have: $f(x_n,y_n) = 1 ,\space \forall n$. Thus: $\displaystyle \lim_{n \to \infty} f(x_n,y_n) = 1 \neq 0 = f(0,0)$. So $f$ is not continuous at $(0,0)$.

calculating partial derivatives at $(0,0)$

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