Warning : it's slightly stupid for a question, but i just want to assure it.
Given question :
$$\lim_{z \to 2i} \frac{z^2+4}{z-2i}=4i$$
My proof :
$$\forall \varepsilon\gt 0,\, \exists \delta\gt 0, \ni \, 0\lt \left|z-2i \right|\lt \delta \Rightarrow \left|\frac{z^2+4}{z-2i} - 4i \right|\lt \varepsilon$$ Sketch work : For $z\ne 2i$
$\begin{align} \left|\frac{z^2+4}{z-2i} - 4i \right| &=\left|\frac{z^2-4iz-4}{z-2i}\right| \\ &=\left|\frac{(z-2i)(z-2i)}{z-2i}\right| \\ &=\left|z-2i \right| \\ &\lt \varepsilon \\ \text{Set } \left|z-2i\right|\lt \varepsilon \text{ to be equal to } \delta, \text{ we conclude that } \varepsilon = \delta \end{align}$
Formal Proof:
Given $\varepsilon \gt 0$, choose $\delta = \varepsilon$ such that for $z\ne 2i$ we have the following:
Assume $0\lt \left|z-2i\right|\lt \delta$ $\begin{align} \Rightarrow\left|\frac{z^2+4}{z-2i} - 4i\right| &=\left|\frac{z^2-4iz-4}{z-2i}\right| \\ &=\left|\frac{(z-2i)(z-2i)}{z-2i}\right| \\ &=|z-2i|\\ &\lt \delta \\ &= \varepsilon \end{align}$
It's weird to get $\delta = \varepsilon$ when usually i got $\delta$ is multiple of $\varepsilon$. Is there anything wrong with my proof? If it isn't, could i conclude that "if we have a rational function with the numerator is a quadratic function that has two identical roots and the denominator is its factor and the difference of $z$ and $z_0$ is equal to the denominator, then if we wanna proof this limit, we have always get $\delta = \varepsilon$" ?
If you don't understand what i mean on the setence above, here is another example that i've tried and i got $\delta = \varepsilon$
$$\lim_{z \to i} \frac{z^2+1}{z-i}=2i$$
In your first example, roots of the numerator are $2i$ and $-2i$ and for second example, roots are $i$ and $-i$.
So correct formulation is:
"If we have a rational function with the numerator is a quadratic function that has two roots which are negatives of each other and the denominator is its factor and the difference of $z$ and $z_0$ is equal to the denominator, then if we wanna proof this limit, we have always get $\delta=\epsilon$."
To verify this solve general example, $$\lim_{z \to a} \frac{z^2- a^2}{z-a}= 2a$$ where $a$ is any complex number.