I was doing an exercise which asks for these partial derivatives, if possible.
\begin{cases} \dfrac{\partial f}{\partial x}(0,0), \dfrac{\partial f}{\partial y}(0,0) \end{cases} Consider this function $f$: \begin{cases} \dfrac{xy}{x^2+y^2} & (x,y) \neq (0,0)\\ 0 & (x,y) = (0,0)\\ \end{cases}
Since I wanna know the derivative at $(0,0)$ is it enough to only calculate the limit for the first equation or do I have to check for the second? Can anyone just tell me which limits to do, please?
We will utilize the limit definitions of partial derivatives, as user Peter Foreman notes in the comments. Both subdomains of the function will be utilized in this process. $$f_x(0,0)=\lim\limits_{h\to0}\frac{f(0+h,0)-f(0,0)}{h}=\lim\limits_{h\to0}\frac{f(h,0)-f(0,0)}{h}$$ We know from the function definitions that $f(0,0)=0$. $$\lim\limits_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim\limits_{h\to0}\frac{f(h,0)}{h}$$ Now, as was discussed in the comments, the limit as $h\to0$ does not have $h$ ever equal $0$, only approach it. As such, $h$ will always be a nonzero real number. Consequently, we will have to follow the $(x,y)\ne(0,0)$ subdomain to finish solving the limit. $$\lim\limits_{h\to0}\frac{f(h,0)}{h}=\lim\limits_{h\to0}\frac{\frac{(h)(0)}{h^2+0^2}}{h}=\lim\limits_{h\to0}\frac{\frac{0}{h^2}}{h}=\lim\limits_{h\to0}\frac{0}{h}=0$$ Next, let us find $f_y(0,0)$ utilizing the same method as with $f_x(0,0)$. $$f_y(0,0)=\lim\limits_{h\to0}\frac{f(0,0+h)-f(0,0)}{h}=\lim\limits_{h\to0}\frac{f(0,h)}{h}=\lim\limits_{h\to0}\frac{\frac{(0)(h)}{0^2+h^2}}{h}$$ Clearly, we will come to the same answer as before. $$\lim\limits_{h\to0}\frac{\frac{(0)(h)}{0^2+h^2}}{h}=\lim\limits_{h\to0}\frac{\frac{0}{h^2}}{h}=\lim\limits_{h\to0}\frac{0}{h}=0$$ Thus, $f_x(0,0)=0$ and $f_y(0,0)=0$ as well.