Say we are to evaluate the following complex-valued limit $(1)$ via the $\varepsilon$$-$$\delta$ definition.
$$\lim\limits_{z\to i}\left(\frac{iz^3 - 1}{z + i}\right), \ \ \ \ z \in \Bbb{C},\tag1$$
I have tried different techniques, such as factorisation, multiplying by complex conjugate, various inequalities... Still, I always hit a dead end.
- How can I prove that this limit is $0$ via the epsilon$-$delta definition?
Let us first rearrage $(1)$ a bit to a nicer form for our purposes.
$$\lim\limits_{z\to i}\left(\frac{iz^3 - 1}{z + i}\right) = \lim\limits_{z\to i}\left[\frac{i\left(z^3 - 1/i\right)}{z + i}\right] = \lim\limits_{z\to i}\left[\frac{i\left(z^3 + i\right)}{z + i}\right]\tag{1'}$$
First find an estimate of what the limit is. From $(1')$ with direct substitution,
$$\lim\limits_{z\to i}\left[\frac{i\left(z^3 - 1/i\right)}{z + i}\right] = \frac{i\left(-i + i\right)}{2i} = 0. \tag2$$
What remains is to prove that $0$ actually is the limit of $(1)$.
$$\left|\frac{i\left(z^3 + i\right)}{z + i} - 0\right| = \left|\frac{z^3 + i}{z + i}\right| = \left|\frac{(z - i)\left(z^2 + iz - 1\right)}{z + i}\right| \leq \left|\frac{|z - i|\left(|z|^2 + |z| + 1\right)}{z + i}\right|\tag 3$$
$$\frac{|z - i|\left(|z|^2 + |z| + 1\right)}{|z + i|} < \delta\frac{|z|^2 + |z| + 1}{|z + i|}\tag{3'}$$
Choose $\delta < 1$; this guarantees an open punctured unit disc centered at $i$. In this region, $|z| < 2$ from which
$$|z|^2 + |z| + 1 < 7.$$
In contrast, we need a lower bound for $|z + i|$. This is a bit trickier. Notice that in this open region given by $\delta < 1$ all complex numbers have a positive imaginary part. That is, $\Im{(z)} = b > 0$.
$$|z + i| = |a + bi + i| = |a + (b + 1)i| = \sqrt{a^2 + (b + 1)^2} > 1\tag4$$
In conclusion,
$$0 < |z - i| < \delta \implies \left|\frac{i\left(z^3 + i\right)}{z + i} - 0\right| < 7\delta < \varepsilon.\tag5$$
Choosing
$$\delta = \min\left\{1,\frac{\varepsilon}{7}\right\}\tag6$$
completes the proof.