Show from the definition of the derivative that $f(z) = Re(z)$ is not complex differentiable at any point.
Easy with the Cauchy-Riemann equations, but I need to do it a different way. Here's my solution, which is convincing enough for me, but there seems to be a subtlety at the end…
Let $z=x+iy$. Then by definition, $f'(z)=\lim\limits_{h \to 0}\frac{f(z+h)-f(z)}{h}$. If this limit exists, it must exist independently of the direction in which $h$ approaches $0$.
Let $h\to0$ along the real axis. Then $z+h=(x+h)+iy$, so $f(z+h)=x+h$. Thus we have $$f'(z)=\lim\limits_{h \to 0}\frac{x+h-x}{h}=1$$
Now let $h\to0$ along the imaginary axis. Then $z+h=x+i(y+h)$, so $f(z+h)=x$. Hence $$f'(z)=\lim\limits_{h \to 0}\frac{x-x}{h}=\frac{0}{0}$$ Here is where I'm unsure of myself. Originally I said $\frac{0}{0}$ is undefined, and so is certainly not equal to $1$, but I thought about L'hospital's rule, and it seems more appropriate to call $\frac{0}{0}$ "indeterminate".
Anyway, is it alright to claim that $1 \neq \frac{0}{0}$? And if so, is my reasoning correct for this proof?
Best Answer
The last limit is $0$, it is not indeterminate.
An indeterminate form arise when you have two functions that tend to $0$ and you take their quotient; I think it is important to understand that what this "indeterminate" form tells you is that you don't have enough information to understand if the limit is $0$, $\infty$, or any other constant.
Indeterminate forms are just a lack of information; in fact, through manipulations (hopital, taylor expansion, etc) you can usually find the result; simply, it is not apparent because the $0/0$ expression tells you nothing about what really happens.
In this case though you have $0/h$. Now, for every $h$, this number is clearly equal to $0$; and $$\lim_{h \to 0} 0 = 0$$
The rest of the proof is correct