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Given $|z|<1$ with $z \in \mathbb{C}$ prove $\lim_{n \to \infty} z^n=0$.

My attempt given $\epsilon >0$ we wish to show that there exists $\delta >0$ such that $n> \delta$ means $|z^n-0| < \epsilon$.

For $n \in \mathbb{N}$. We have $|z^n|=|z|^n$. We wish $|z|^n < \epsilon$. This means,

$$n \ln |z| < \ln \epsilon$$

But because $\ln |z|<\ln 1=0$ dividing both sides by $\ln |z|$ gives:

$n> \frac{\ln \epsilon}{\ln |z|}$

So that's what we choose for $\delta$.

My question is is how can we get away from the assumption that $n \in \mathbb{N}$. Or is this correct even with the assumption?

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    When you write $\lim_{n\to \infty}$, it's usually implicit that $n$ is restricted to integers / natural numbers, just because it's $n$ instead of $x$ or $a$. It's not a good thing, really, but it is what it is. You can't be completely certain without reading the mind of whoever wrote it. – 2017-02-18

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The issue here is that $z^n$ is multivalued when $n>0$ real is not an integer. The same argument though shows that whichever possibility you choose will still have absolute value approaching $0$. This misses the point, though; there's no reason to generalize to $n\notin \mathbb N$ for this problem.