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:


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:


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


1 Answer 1


You can also proceed as follows as to the second term.

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*}

and we are done.

Hopefully this helps !


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