Let a function $f$ be defined at all points $z$ in some deleted neighborhood of $z_0.$ The statement that the limit of $f (z)$ as $z$ approaches $z_0$ is a number $w_0,$ or that \begin{eqnarray}\label{limit001} \lim_{z\rightarrow z_0}f(z)=w_0 \end{eqnarray} means that the point $w = f (z)$ can be made arbitrarily close to $w_0$ if we choose the point $z$ close enough to $z_0$ but distinct from it.

Formally, the expression (\ref{limit001}) means that for every (sufficiently small) $\varepsilon>0,$ there is a $\delta>0$ such that

\begin{eqnarray}\label{limitcondition} \left|f (z) - w_0\right| \lt \varepsilon \quad \text{whenever}\quad 0 \lt \left|z - z_0\right| \lt \delta \end{eqnarray}

Under the mapping $w = f (z),$ all points interior to the circle $|z - z_0| = \delta$ with $z_0$ deleted are mapped to points interior to the circle $|w - w_0| = \varepsilon,$ see Figure 1. The limit will exist only in the case when $z$ approaches $z_0$ (that is, $z \rightarrow z_0$) in an arbitrary direction; then this implies that $w \rightarrow w_0.$

Geometric interpretation of limit.

Limits are unique if they exist.
Suppose that $\lim_{z\rightarrow z_0}f(z)=w_0$ and $\lim_{z\rightarrow z_0}f(z)=w_1$ with $w_0\neq w_1.$ Let $2\varepsilon =|w_0-w_1|,$ so that $\varepsilon >0.$ There is a $\delta>0$ such that $0\lt|z-z_0|\lt\delta$ implies that $|f(z)-w_0|\lt\varepsilon$ and $|f(z)-w_1|\lt\varepsilon.$ Choose a point $z$ different to $z_0$ (because $f$ is defined in a deleted neighborhood of $z_0$). Then, using the triangle inequality, $$|w_0-w_1|\leq |w_0-f(z)|+|f(z)-w_1|\lt2\varepsilon.$$ But this is a contradiction. Thus $w_0=w_1.$ $\quad \blacksquare$

Example 1: Show that $$\lim_{z\rightarrow z_0}z^2=z_0^2.$$

Discussion: Let's take $\delta = 1.$ So $0\lt\left| z-z_0\right|\lt\delta=1$ implies that

\begin{eqnarray*} \left| z^2-z_0^2\right|&\lt&\left| z-z_0\right|\left| z+z_0\right|\\&\lt&\left| z-z_0+2z_0\right|\\&\lt& \left(\left| z-z_0\right|+2\left|z_0\right|\right)\\ &\lt&\left(1+2\left|z_0\right|\right). \end{eqnarray*}
Now, if $\delta=\dfrac{\varepsilon}{1+2\left|z_0\right|},$ then $0\lt\left| z-z_0\right|\lt\delta$ implies that $$\left| z^2-z_0^2\right|\lt \left(1+2\left|z_0\right|\right)\lt\varepsilon.$$ This means that $$\left| z^2-z_0^2\right|\lt\varepsilon \quad\text{whenever}\quad 0\lt\left| z-z_0\right|\lt\delta$$ where $\delta=\min\left\{1,\dfrac{\varepsilon}{1+2\left|z_0\right|}\right\}.$

Now we can proceed to write our formal proof.

Proof: Let $\varepsilon>0.$ Choose $\delta=\min\left\{1,\dfrac{\varepsilon}{1+2\left|z_0\right|}\right\}$ such that $0\lt\left| z-z_0\right|\lt\delta.$ Therefore $$ \left| z^2-z_0^2\right|\lt\varepsilon. \quad \blacksquare $$

There is a connection between the limits of complex functions $f(z)$ and the limits of real-valued functions of two real values $g(x,y).$ The latter type are studied in calculus and we can use their definition and properties, like the following:

Consider $f(z)=u(x,y)+iv(x,y)$ and $z_0=x_0+iy_0,$ $w_0=u_0+iv_0.$ Then \begin{eqnarray}\label{limitpart01} \lim_{z\rightarrow z_0}f(z)=w_0 \end{eqnarray} if and only if \begin{eqnarray}\label{limitpart02} \lim_{\left(x,y\right) \rightarrow \left(x_0,y_0\right)}u\left(x,y\right)=u_0\quad \text{and}\quad \lim_{\left(x,y\right) \rightarrow \left(x_0,y_0\right)}v\left(x,y\right)=v_0 \end{eqnarray}

First, let us assume that limit (\ref{limitpart01}) holds. Thus, for each positive number $\varepsilon,$ there is a positive number $\delta$ such that \begin{eqnarray}\label{limpart05} |(u + iv) -(u_0 + iv_0)| \lt \varepsilon \end{eqnarray} whenever \begin{eqnarray}\label{limpart06} 0 \lt |(x + iy) - (x_0 + iy_0)| \lt \delta \end{eqnarray} But

\begin{eqnarray*} |u-u_0|&=&|(u-u_0) + i(v -v_0)|\;=\;|(u + iv) - (u_0 + iv_0)|,\\ |v-v_0|&=&|(u-u_0) + i(v -v_0)|\;=\;|(u + iv) - (u_0 + iv_0)| \end{eqnarray*}
Hence it follows from inequalities (\ref{limpart05}) and (\ref{limpart06}) that
$$|u-u_0|\lt\varepsilon\quad \text{and}\quad |v-v_0|\lt\varepsilon$$
which proves that limits (\ref{limitpart02}) hold.

Now, we assume that limits (\ref{limitpart02}) hold in order to obtain limit (\ref{limitpart01}). Limits (\ref{limitpart02}) tell us that for each positive number $\varepsilon,$ there exist positive numbers $\delta_1$ and $\delta_2$ such that

\begin{eqnarray}\label{limpart03} |u-u_0|\lt \frac{\varepsilon}{2}\quad\text{whenever}\quad 0\lt\sqrt{(x-x_0)^2+(y-y_0)^2}\lt\delta_1 \end{eqnarray}
\begin{eqnarray}\label{limpart04} |v - v_0| \lt \frac{\varepsilon}{2}\quad\text{whenever}\quad 0\lt\sqrt{(x-x_0)^2+(y-y_0)^2}\lt\delta_2 \end{eqnarray}
Let $\delta$ be any positive number smaller than $\delta_1$ and $\delta_2.$ Since
$$|(u + iv) - (u_0 + iv_0)| = |(u - u_0) + i(v - v_0)| \leq |u - u_0| + |v - v_0|$$
$$\sqrt{(x-x_0)^2+(y-y_0)^2}=|(x-x_0)+i(y-y_0)|\leq |(x+iy)-(x_0+iy_0)|$$
it follows from statements (\ref{limpart03}) and (\ref{limpart04}) that
$$|(u + iv) -(u_0 + iv_0)| \lt\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$$
$$0 \lt |(x + iy) - (x_0 + iy_0)| \lt \delta.$$
That is, limit (\ref{limitpart01}) holds, and this completes the proof of the theorem. $\blacksquare$

Example 2: Show that $$\lim_{z\rightarrow 0}\dfrac{z}{\overline{z}}$$ does not exist.

Solution: Suppose that the limit exists. Thus we can calculate it by letting the point $z = x+iy$ approach the origin in any manner. However, when $z = x+i0$ is a nonzero point on the real axis $$f(z)=\frac{x+i0}{x-i0}=1;$$ and when $z=0+iy$ is a nonzero point on the imaginary axis, $$f(z)=\frac{0+iy}{0-iy}=-1.$$

Thus, if we let $z$ approach the origin along the real axis, we found that the limit is $1.$ On the other hand, if we approach along the imaginary axis we found the limit $-1,$ which is a contradiction because a limit is unique. Therefore we must conclude that limit $$\lim_{z\rightarrow 0}\dfrac{z}{\overline{z}}$$ does not exist.

In addition to computing specific limits, Theorem 2 is also an important theoretical tool that allows us to derive many properties of complex limits from properties of real limits. The following theorem gives an example of this procedure.

Suppose that \begin{eqnarray} \lim_{z\rightarrow z_0}f(z)=w_1\quad \text{and}\quad \lim_{z\rightarrow z_0}g(z)=w_2\label{limithypo} \end{eqnarray} Then \begin{eqnarray} \lim_{z\rightarrow z_0}\big[c\cdot f(z)\big] &=& c\cdot \lim_{z\rightarrow z_0} f(z) \text{ with } c\in\C, \label{limitconst}\\ \lim_{z\rightarrow z_0}\big[f(z)+g(z)\big]=w_1+w_2,\label{limitsum}\\ \lim_{z\rightarrow z_0}\big[f(z)g(z)\big]=w_1w_2;\label{limitmult} \end{eqnarray} and, if $w_2\neq 0,$ \begin{eqnarray} \lim_{z\rightarrow z_0}\frac{f(z)}{g(z)}=\frac{w_1}{w_2}. \end{eqnarray}


For proving property (\ref{limitsum}), we must show that for any $\varepsilon >0$ we can find $\delta >0$ such that

\begin{eqnarray*} \left| \left(f(z)+g(z)\right)-\left(w_1+w_2\right)\right|\lt\varepsilon \quad \text{when}\quad 0\lt\left| z-z_0\right|\lt\delta \end{eqnarray*}
By hypothesis, given $\varepsilon >0$ we can find $\delta_1 >0$ and $\delta_2>0$ such that
\begin{eqnarray*} \left| f(z)-w_1\right|\lt\varepsilon/2 \quad \text{when}\quad 0\lt\left| z-z_0\right|\lt\delta_1 \\ \left| g(z)-w_2\right|\lt\varepsilon/2 \quad \text{when}\quad 0\lt\left| z-z_0\right|\lt\delta_2 \end{eqnarray*}
Then we have
\begin{eqnarray*} \left| \left(f(z)+g(z)\right)-\left(w_1+w_2\right)\right| &=&\left| \left(f(z)-w_1\right)+\left(g(z)+w_2\right)\right| \\ &\leq &\left| f(z)-w_1\right|+\left| g(z)-w_2\right|\\ &\leq & \varepsilon/2+\varepsilon/2=\varepsilon \end{eqnarray*}
\begin{eqnarray*} \left| \left(f(z)+g(z)\right)-\left(w_1+w_2\right)\right|\lt\varepsilon \quad \text{when}\quad 0\lt\left| z-z_0\right|\lt\delta \end{eqnarray*}
where $\delta$ is chosen as the smaller of $\delta_1$ and $\delta_2.$

Now we will prove property (\ref{limitmult}). First, using the triangle inequality, we write

\begin{eqnarray*} \left|f(z)g(z) - w_1w_2\right|& \leq & \left|f(z)g(z) - f(z)w_2\right| + \left|f(z)w_2 - w_1w_2\right|\\ &=& \left|f(z)\right|\left|g(z) - w_2\right|+ \left|f(z) - w_1\right|\left|w_2\right| \end{eqnarray*}
To estimate each term, we choose $\delta_1\gt 0$ so that $0\lt \left|z-z_0\right|\lt \delta_1$ implies $\left|f(z)- w_1\right|\lt 1,$ and thus $\left|f(z)\right|\lt \left|w_1\right|+1,$ since $$\left|f(z) - w_1\right|\geq \left|f(z)\right|- \left| w_1\right|.$$ Given $\varepsilon\gt 0,$ set $\delta_2$ and $\delta_3$ such that $0\lt \left|z-z_0\right|\lt \delta_2$ implies $$\left|f(z)-w_1\right|\lt \frac{\varepsilon}{2\left(|w_2|+1\right)}$$ and $0\lt \left|z-z_0\right|\lt \delta_3$ implies $$\left|g(z)-w_2\right|\lt \frac{\varepsilon}{2\left(|w_1|+1\right)}.$$ Thus we can choose $\delta = \min\left\{\delta_1,\delta_2,\delta_3\right\}.$ If $0\lt \left|z-z_0\right|\lt \delta,$ then
\begin{eqnarray*} \left|f(z)g(z) - w_1w_2\right|& \leq & \left|f(z)\right|\left|g(z) - w_2\right|+ \left|f(z) - w_1\right|\left|w_2\right|\\ &\lt & \frac{\varepsilon}{2\left(|w_1|+1\right)}\left|f(z)\right| + \frac{\varepsilon}{2\left(|w_2|+1\right)}\left|w_2\right|\\ &\lt & \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{eqnarray*}
Therefore, $\lim_{z\rightarrow z_0}\big[f(z)g(z)\big]=w_1w_2$ is proved.

Properties (\ref{limitconst}) and (\ref{limitsum}) are left as an excersie. $\blacksquare$

Riemann Sphere