In the Caucy-Riemann equations, for an open set $\ U$$\subseteq$$\mathbb C$, $\ f:U\to\mathbb C $ is a function of a complex variable. $\ f$ is differentiable at $c=a +ib\in U$. Let $$ f(x+iy) = u(x,y) +iv(x,y)$$
Then, the partial derivatives $$\frac {\partial u}{\partial x} , \frac {\partial u}{\partial y},\frac {\partial v}{\partial x}, \frac {\partial v}{\partial y} $$ exists at (a,b)
My question is 'why do the partial derivatives exist?'. In the proof of the Cauchy-Riemann equations, we consider $h\to0$ along the real axis, and so $$\frac {f(a+ib+h)-f(a+ib)}{h} = \frac {u(a+h,b)-u(a,b)}{h}+i\frac{v(a+h,b)-v(a,b)}{h}$$ So I learned as $h\to0$ the right hand side goes to $\frac{\partial u}{\partial x}$ at $(a,b)$ and $i\frac {\partial v}{\partial x}$ at $(a,b)$, i.e. the partial derivatives exist at $(a,b)$. But I thought we cannot separate the right hand side as $\frac{\partial u}{\partial x}$ at $(a,b)$ and $i\frac {\partial v}{\partial x}$ at $(a,b)$ becuase we don't know if the limits of each of those exist yet. How do we know the partial derivatives exist? Thank you so much!
Since $f$ is differentiable at $c$, those partial derivatives exist. Since you know that the limit$$\lim_{h\to0}\frac{f(a+bi+h)-f(a+bi)}h$$exists and it is equal to a number $\alpha+\beta i$, then, in particular,$$\lim_{h\to0,\ h\in\mathbb R}\frac{f(a+bi+h)-f(a+bi)}h=\alpha+\beta i.$$But then$$\lim_{h\to0,\ h\in\mathbb R}\frac{u(a+h,b)-u(a,b)}h=\alpha\text{ and }\lim_{h\to0,\ h\in\mathbb R}\frac{v(a+h,b)-v(a,b)}h=\beta,$$which means that the partial derivatives $\frac{\partial u}{\partial x}(a,b)$ and $\frac{\partial v}{\partial x}(a,b)$ exist (and are equal to $\alpha$ and to $\beta$ respectively).
The same argument proves that the paartial derivatives $\frac{\partial u}{\partial y}(a,b)$ and $\frac{\partial v}{\partial y}(a,b)$ exist.