# Question about limit definition of partial derivative

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?

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Unless I am not missing something, the two formulas are exactly the same. The only difference is notation for the point where you evaluate the partial derivative. In one of the it is denoted $(x,y)$, in the other one $(x_0,y_0)$. – Martin Sleziak May 18 '14 at 5:14

$y_0$ 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.
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 $(0,0)$.