Remember the definition of limit,
$$\forall \epsilon\; \exists \delta : |x-a|<\delta \rightarrow |f(x) - L|<\epsilon $$
Now if $L$ is not the limit negate the definition, that is
$$ \exists \epsilon\; \forall \delta: \;\; \exists x \text{ s.t. } |x-a| <\delta\text{ and } |f(x) - L| \geq \epsilon $$
In your case, as some people already mentioned take $\epsilon =1/2$, now for ever $\delta$ we need to find an $x$ such that $|x-4|<\delta$ and $|9-x-4|\geq 1/2$. Finding such $x$ isn't that difficult, just take $0<r<\min\{\delta, 1/2\} $ and consider $x=4+r$. To check we substitute in the inequalities:
$$|x-4|=|r|<\delta$$
And
$$|9-x-4|=|1-r|>1/2= \epsilon $$
(sorry for the awful format, I'm on my phone)
The point of the definition of limit is to capture the idea that we can force the values of $f$ to be close enough to $L$, provided only that the values of $x$ be sufficiently near $a$: if you tell me how close you want $f(x)$ to be to $L$, I can guarantee this outcome by telling you how close $x$ should be to $a$.
Reversing the logical dependency between $\epsilon$ and $\delta$ makes the logical dependency run the "wrong" way: you are saying that you will tell me how close you want $x$ to be to $a$, and then I will be forced to tell you how close I can guarantee $f(x)$ to be to $L$.
It might seem like this would be good enough, but it doesn't work. It seems to be the same because you may be thinking of the limit as saying "the closer you get to $a$, the closer the values will get to $L$"; but saying the limit is $L$ is more than that: it says that values get arbitrarily close to $L$, and that all values get close to $L$ near $a$, not just some.
If you are allowed to pick the value of $\epsilon$, then you are not guaranteeing that the values get arbitrarily close to $L$, just that they get "sufficiently" close to $L$.
So for example, you want the limit of $f(x)$ to approach at most one thing, not two or more. But say that $f(x)$ always takes values between $-1$ and $1$, as $f(x)=\sin(x)$ does. If I take $L$ to be any value between $-1$ and $1$, and then let $\epsilon=3$, then regardless of what your $\delta$ is, we will indeed satisfy that $|f(x)-L|\lt \epsilon$ whenever $|x-a|<\delta$. So every number between $-1$ and $1$ is a limit. And worse, any number is a limit: if you give me $L=10$, then provided I let $\epsilon>11$, every value of $f(x)$ will be within $\epsilon$ of $L$.
That means this definition doesn't really capture the notion we want the definition of limit to capture.
Remember: to convince me that the limit is $L$, you challenge me to fall within an arbitrarily thin horizontal band around $L$. The challenge lies in how thin the horizontal band it; if you let me pick how thin that band is, then I can make it really fat and have absolutely no challenge at all.
Best Answer
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.