I have $$\lim_{z \to i} \frac {iz^3-1}{z+i}=0$$
To prove this I am trying to use the epsilon-delta definition. By saying that for any $\delta >0$ and any $\varepsilon >0$ then:
$$0<|z-i|<\delta \quad \mathrm {whenever} \quad \left|\frac {iz^3-1}{z+i}\right|< \varepsilon$$
I factor the implication on the right to get:
$$\left|\frac {(z-i)(z^2+iz-1)}{z+i}\right| < \varepsilon$$
I know that I need to "solve" for $z-i$ and in order to do so I need to see how the other terms on the LHS behave when in a close neighborhood of $z-i$. Therefore, I pick an arbitrary value such that $|z-i| < \delta \le 1$.
This is how I have seen other people do it with real limits, however, I get stuck here because when I solve for $z$ with use of the triangle inequality, I get that $|z| \le 0$ which is obviously not true. I can see that I could choose another bound, but is this not supposed to work for any arbitrary value? Also, if it works, how should I proceed?
1 Answer
$\begingroup$If we asume $|z-i| < \delta \le 1$, then $$|z+i| = |-2i - (z-i)| \ge |2i| - |z-i| > 2 - \delta \ge 1$$ and $$|z^2 + iz - 1| = |(z-i)^2 + 3i(z-i) - 2| \le \delta^2 + 3 \delta + 2 \le 6$$ so $$\Big| \frac{(z-i)(z^2 + iz - 1)}{z+i} \Big| \le 6\delta.$$
Given a fixed $\varepsilon > 0$ (small enough) you can let $\delta = \varepsilon / 6.$
$\endgroup$
