Question about limit definition of partial derivative

Question

I've seen it written two different ways:

$$\frac{\partial f}{\partial x} = \lim\limits_{h \rightarrow 0} \frac{f(x + h, y) - f(x,y)}{h}$$

and

$$\frac{\partial f}{\partial x} = \lim\limits_{h \rightarrow 0} \frac{f(x_0 + h, y_0) - f(x_0,y_0)}{h}$$

where the latter evaluates the function at the respective point before plugging it into the definition of the limit. For example, the function $f(x,y) = \begin{cases} \frac{x^2 y^4}{x^4 + 6y^8}, & \text{if }(x,y) \neq (0,0) \\ 0, & \text{if }(x,y) = (0,0) \end{cases}$

I want to determine if $\frac{\partial f}{\partial x}$ exists at $(0,0)$.

Using the second limit definition would make showing the existence of $\frac{\partial f}{\partial x}$ considerably easier, since $y_0$ makes the first term in the limit $0$, and $f(x_0,y_0)$ is defined to be $0$.

But using the first definition, we have to evaluate:

$$\frac{(x+h)^2 y^4}{(x+h)^4 + 6y^8} - \frac{x^2 y^4}{x^4 + 6y^8}$$

I'm hoping the "real" or at least usable definition is the second one, but which one is the one we're supposed to use in practice to be technically/mathematically correct?

Answer 1

$y_0$ does not mean "set $y$ to $0$", it just indicates a particular value of $y$. The two formulas suggest finding the partial derivative for any general point $(x,y)$ and finding the partial derivative for a specific point $(x_0,y_0)$ [but not necessarily $(0,0)$]. Of course that's just interpretation though.

Answer 2

If the limit exists, then the limit should be independent of the path taken to $(0,0)$. Let $y = m\sqrt{x}$ where $m \in \mathbb{R}$. Then $$\lim_{(x,y) \to (0,0)}\frac{x^2 y^4}{x^4 + 6 y^8} = \lim_{x \to 0} \frac{m^4 x^4}{x^4 + m^8 x^4} = \frac{m^4}{1 + 6m^8}.$$ The limit clearly changes depending on the value $m$. So $f(x,y)$ is not differentiable at $(0,0)$.

This method can only be used to disprove that the limit exists.