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Find the limit, or give an argument of why it doesn't exist

$\lim_{z \to i} \dfrac {iz^3-1}{z+i}$

After many failed attempts, I looked at the answers and saw the limit is zero. However, I still can't solve it. I think the problem is meant to be solved using $\epsilon- \delta$, but I guess we can use other ways if really necessary.

I know I have to start at $\left|\dfrac {iz^3-1}{z+i} -0\right| < \epsilon$ and do illegal manipulations (make the LHS bigger) and arrive at $|z - i| < h(\epsilon)$. I proceeded:

$\left|\dfrac {iz^3-1}{z+i} -0\right| = \left|\dfrac {i(z^3+i)}{z+i}\right| = |i|\left|\dfrac { z^3 - i^3}{z+i}\right| = \left|\dfrac { (z - i)(z^2 + zi - 1)}{z+i}\right| $

The only option I can see is to try to find an upper bound for $\left|\dfrac {z^2 + zi - 1}{z+i}\right|$ on some disk centered at $i$, but I'm really not sure how to do this.

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    Do you have some result about limits of quotients $p(z)/q(z)$ in your course? – 2017-02-10
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    @ThibautDumont Haha that's a very good point, I was convinced that the book suggested we try this exercise before it introduced the limit rules. However, I just checked and the book doesn't give this suggestion. – 2017-02-10
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    @ThibautDumont I will try to solve it with the quotient rule now, but I think it would still be nice if somebody showed a way to do it using brute force. – 2017-02-10
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    Hint: the function is continuous, and has no singularity at $z=i\,$, so the limit is simply $f(i)$. – 2017-02-10
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    @dxiv I am very new to Complex Analysis, would you mind briefly informing me of which types of functions from $\mathbb{C}$ to $\mathbb{C}$ are continuous? For example, are all polynomial and rational functions continuous barring singularities? – 2017-02-10
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    @Ovi Polynomial functions, yes. Rational functions, also yes except at points where the denominator vanishes. This follows from the more general properties that sums and products of continuous functions are themselves continuous, and ratios thereof are continuous except at points where the denominator vanishes. – 2017-02-10
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    @dxiv Thank you. Last question; what about functions with non-integer exponents? Are they generally continuous? – 2017-02-10
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    @Ovi That's a broader, more loaded question. Loosely speaking, those functions can not be defined to be continuous on the *entire* $\mathbb{C}$ because they are [multi-valued](http://mathworld.wolfram.com/MultivaluedFunction.html) in the general case, but are continuous on subsets of $\mathbb{C}$ which exclude some chosen [branch cuts](http://mathworld.wolfram.com/BranchCut.html). – 2017-02-10
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    @Ovi That sorta depends. Take $z^{1/2}\, \displaystyle{\operatorname*{ =}^{\text{sorta}} \sqrt{z}}$ and observe that branch cuts cause problems – 2017-02-10
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    @dxiv Thank you very much – 2017-02-10
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    @BrevanEllefsen Thank you very much – 2017-02-10
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    Why not derive some bounds geometrically? For starters: $$\left|\frac ab\right|=\frac{|a|}{|b|}$$and from there, show $3>|z+i|>1$ for $|z-i|<1$ and $$|z^2+iz-1|=|(z+i)^2+i(z-i)-1|\le|z+i|^2+|z-i|+1<3^2+1+1=11$$ – 2017-03-04

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