How is $u_{xy}=u_{yx}$ trivially?

Question

Let $f(z)=u(x,y)+iv(x,y)$ be a complex mapping. My book "Complex analysis" by Brown and Churchill says that if $u_x$ and $u_y$ are continuous, then $u_{xy}=u_{yx}$.

I have formed a sketchy proof, but I don't know where the "$u_x,u_y$ are continuous" part comes in.

$$u_{xy}=\displaystyle{\lim\limits_{\Delta y\to 0}\frac{\left[\lim\limits_{\Delta x\to 0}\frac{u(y+\Delta y,x+\Delta x)-u(y+\Delta y,x)}{\Delta x}-\lim\limits_{\Delta x\to 0}\frac{u(y,x+\Delta x)-u(y,x)}{\Delta x}\right]}{\Delta y}}$$ Similarly, $$u_{yx}=\displaystyle{\lim\limits_{\Delta x\to 0}\frac{\left[\lim\limits_{\Delta y\to 0}\frac{u(y+\Delta y,x+\Delta x)-u(y,x+\Delta x)}{\Delta y}-\lim\limits_{\Delta y\to 0}\frac{u(y+\Delta y,x)-u(y,x)}{\Delta y}\right]}{\Delta x}}$$

On simplification, both of them turn out to be equal to $$\lim\limits_{\Delta x,\Delta y\to 0}\frac{u(x+\Delta x,y+\Delta y)-u(x+\Delta x,y)-u(x,y+\Delta y)+u(x,y)}{\Delta x.\Delta y}$$

1. As both of them essentially boil down to the same expression, they have to be equal. Is this reasoning right?
2. Where is the "partial derivatives are continuous" condition required?

Thanks in advance!

Answer 1

No, your argument is not correct. In place of $$ \lim_{\Delta x, \Delta y \to 0} $$ you get $$ \lim_{\Delta x\to 0}\;\lim_{\Delta y \to 0} $$ in one case, and $$ \lim_{\Delta y\to 0}\;\lim_{\Delta x \to 0} $$ in the other.