I was reading through the book "Complex Analysis" by John M. Howie. On page 54, he goes through a proof of the partial converse of the Cauchy Riemann equations – that is, given an open neighbourhood where the complex function has continuous partial derivatives and satisfies the Cauchy Riemann equations, the derivative of the complex function exists at every point within this neighbourhood. He goes through a proof, but at one stage essentially concludes
$${\lim_{l\rightarrow 0}\frac{1}{l}\left(\epsilon_1h + \epsilon_2 k + i\epsilon_3 h + i\epsilon_4 k\right)=0}$$
where ${l=h+ik}$, and ${\epsilon_1 = \frac{\partial u}{\partial x}(a+\text{a bit},b+\text{a bit})-\frac{\partial u}{\partial x}(a,b)}$ (where the "a bit" is just some junk involving ${h,k}$) and the other epsilons are defined similarly (in terms of the other partial derivatives). Because of the continuity of the partial derivatives, these epsilons go to $0$ as ${(h,k) \rightarrow (0,0)}$, to which I agree. But I'm not sure how he concludes the limit above is $0$? Am I missing something? If it helps, I can post a picture of the proof in question if I'm allowed to do so
Best Answer
Since $l=h+ik$, we have $|h|\leq |l|$. So $|\frac hl|$ is bounded above. Thus $$\frac{\epsilon_1 h}{l}\to 0$$ as $l\to 0$ since $\epsilon_1\to 0$ whenever $l\to 0$. Similarly, you can conclude for the other terms.