Let $f:\mathbb C\to\mathbb C$ such that the limit $\lim_{z\to 0}f(z)=z_0\in\mathbb C$. I would like to show that this implies
$(i)$. $\lim_{r\to 0}f(r)=z_0$ when "treating $r$ to be real-valued".
$(ii)$. $\lim_{r\to 0}f(ir)=z_0$ when "treating $r$ to be real valued so that $ir$ is purely imaginary."
I saw them on every complex analysis textbook, but I cannot even formulate a formal way of stating "treating $r$ real valued". What kind of result is using here? And also, do these two conditions give rise to a sufficient and necessary condition for the complex-valued limit to exist?
In particular the proof of Cauchy-Riemann equation uses the fact that "by taking $h$ to be purely real and imaginary the limit $\lim_{h\to 0}\frac{f(z_0+h)-f(z_0)}{h}$ is the same." I cannot find a formal way of stating it.
Following the comment let me start with the definition of the limit of a function:
Let $f$ be as above. $\lim_{z\to 0}f(z)=z_0$ means that for every $\varepsilon>0$ there is a $\delta>0$ such that $|f(z)-z_0|<\varepsilon$ whenever $0<|z|<\delta$. Here absolute value is the Euclidean norm on $\mathbb R^2$. This implies that in particular when $z$ only has real part and $0<|z|<\delta$ then...