# Prove that $\lim\limits_{z\to0}z+\vert{z}\vert^3=0$. Epsilon-Delta proof.

The problem is to prove that: $$\lim\limits_{z\to 0}z+\vert{z}\vert^3=0$$

According to limit rules, we have that $$\left[\lim\limits_{z\to z_0}f(z)+g(z)\right]=\lim\limits_{z\to z_0}f(z)+\lim\limits_{z\to z_0}g(z)$$, and we can therefore evaluate each term separately. In our case we are looking at $$f(z)=z$$ and $$g(z)=\vert{z}\vert^3$$.

For the first term, we have that $$\lim\limits_{z\to 0}f(z)=\lim\limits_{z\to 0}z=0$$. Now, given $$\epsilon>0$$ choose $$\delta=\epsilon$$ and we get:

$$\vert{f(z)-L}\vert=\vert{z-0}\vert<\epsilon\space\space\space\textrm{and}\space\space\space\vert{z-z_0}\vert=\vert{z-0}\vert<\delta$$

As for the second term, we have $$\lim\limits_{z\to 0}g(z)=\lim\limits_{z\to 0}\vert{z}\vert^3=0$$. So, given sufficiently small $$\epsilon>0$$ choose $$\delta=min\{1,\epsilon\}$$. Consider that for $$z\in{D(0,\delta)}$$ any $$\vert{z}\vert^3$$ will be smaller than $$\vert{z}\vert$$, so when $$\vert{z-0}\vert<\delta$$ and $$\vert{z}\vert^3<\epsilon$$, we get:

$$\vert{z}\vert^3<\vert{z}\vert<\delta=\epsilon$$

Firstly, is this correct? And secondly, is this the general way to prove that a complex limit exists or are there any alternatives?

Let $$|z - 0| < \delta_{\varepsilon}$$. Then we have that
\begin{align*} ||z|^{3} - 0| & = |z|^{3} < \delta^{3}_{\varepsilon} := \varepsilon \end{align*}
Then we can conclude that \begin{align*} (\forall\varepsilon > 0)(\exists\delta_{\varepsilon} = \sqrt[3]{\varepsilon} >0)(\forall z\in\mathbb{C})(0 < |z - 0| < \delta_{\varepsilon} \Rightarrow ||z|^{3} - 0| < \varepsilon) \end{align*}