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Can someone help me calculating a partial derivative of the function:

$$f(x,y) = \begin{cases} x \arctan\left[\frac{x}{y}\right] & \text{if } y \neq 0 \\ 0 & \text{if } y = 0 \end{cases}$$

To determinate the partial derivative respect to $x$ in $(0,0)$ $$\lim_{h\to 0} \frac{f(x_0+h,y_0)-f(x_0,y_0)}{h} $$ That becomes $$\lim_{h\to 0} \frac{f(0+h,0)-f(0,0)}{h} $$ so $$\lim_{h\to 0} \frac{h\arctan\left[\frac{h}{0}\right]}{h} $$ However $$\lim_{h\to 0} \arctan\left[\frac{h}{0}\right] $$ is indeterminate. The problem is that the the funtion simply substituting becomes indeterminate. However I know that the value of the partial derivative not calculated with the definition is 0. Someone can help me to find the mistake?

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

$$f_x\left( 0,0 \right) =\lim_{y\rightarrow 0}\lim_{h\rightarrow 0}\arctan \left( \frac{h}{y} \right) $$

Because $arctan0=0$, $f_x(0,y)=0$

$$f_x(0,0)=\lim_{y\to 0}0=0$$

This makes sense because $f(0,y)=0$, so its slope must also be $0$

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