Limits of Complex Functions. Definition, Examples, and Rules

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Why Limits Matter in Complex Analysis

Limits are the bedrock upon which all of complex analysis is built. They enable us to define:

• Continuity: f is continuous at z₀ iff $\lim_{z \to z_0} f(z) = f(z_0)$.
• Derivatives: the derivative of f at z₀ is defined by the limit $f'(z_0) = \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h}$ where h is a complex number approaching 0.
• Integrals: Complex integrals are defined via limits of Riemann sums and many powerful theorems rely on limiting processes.
• Series: An infinite series $\sum_{n=0}^{\infty} a_n z^n$ is understood as the limit of its partial sums: $\sum_{n=0}^{\infty} a_n z^n = \lim_{N \to \infty} \sum_{n=0}^{N} a_n z^n$.

Without a clear understanding of limits, none of these core ideas can be properly grasped.

The Crucial Difference from Real Analysis

In real analysis, when we write $x \to x_0$, the variable x can only approach $x_0$ from the left or from the right along the real line.

In complex analysis, the situation is radically different: z can approach $z_0$ from infinitely many directions – along any curve in the complex plane that ends at $z_0$. For the limit to exist, the value must be the same regardless of the path taken. This path‑independence is a much stronger condition and lies at the heart of many remarkable properties of holomorphic functions.

The Definition of Limit

Before defining the limit of a function, we need the notion of a limit point of a set. This tells us where it is sensible to talk about limits.

Definition. Let $D \subseteq \mathbb{C}$. A point $z_0 \in \mathbb{C}$ (not necessarily in D itself) is a limit point (or accumulation point) of D if every punctured neighborhood of z₀ contains at least one point of D. Formally, $\forall \delta > 0, \quad B'(z_0; \delta) \cap D \neq \emptyset$ where $B'(z_0; \delta) = \{z \in \mathbb{C} : 0 < |z - z_0| < \delta\}$ is the punctured open disk of radius $\delta$ centered at z_0.

• Every point of an open disk is a limit point of that disk.
• Every boundary point of a set is a limit point.
• Isolated points are not limit points.

The point z₀ need not belong to D itself! The definition only requires that we can approach $z_0$ arbitrarily closely through points of D. This is exactly what we need when studying limits: we want to examine the behaviour of f(z) as z gets close to $z_0$ while staying inside the domain of f.

Definition. Let D ⊆ ℂ, $f: D \rarr \Complex$ be a function and z₀ be a limit point of D (so arbitrarily close points of D lie around z₀, though possibly z₀ ∉ D). A complex number L is said to be a limit of the function f as z approaches z₀, written or expressed as $\lim_{z \to z_0} f(z)=L$, if for every epsilon ε > 0, there exists a corresponding delta δ > 0 such that $z \in D \text{ and } 0 < |z - z_0| < \delta \implies |f(z) - L| < \varepsilon$

$\lim_{z \to z_0} f(z) = L \iff \forall \varepsilon > 0, \; \exists \delta > 0 : \left( z \in D \land 0 < |z - z_0| < \delta \right) \Rightarrow |f(z) - L| < \varepsilon$.

In words, for any tolerance ε around L, there's a punctured neighborhood around z₀ whose image under f lies entirely within that tolerance.

If no such L exists, then we say that f(z) does not have a limit as z approaches z₀. This is exactly the same ε–δ formulation we know from real calculus, but now z and L live in the complex plane ℂ and the condition must hold for all z in a two‑dimensional punctured disk intersected with the domain D. Because z can come from any direction, the existence of the limit imposes strong constraints on f.

Equivalently, $\lim_{z \to z_0} f(z) = L \iff \forall \varepsilon > 0, \; \exists \delta > 0 : f\left( D \cap B'(z_0; \delta) \right) \subseteq B(L; \varepsilon)$

All z in the punctured δ-disk around z₀ land in the ε-disk around L.

Why 0 < |z - z₀|? We exclude z = z₀ itself because the limit cares about values near z₀, not at z₀ itself.

The condition 0 < |z - z₀| (equivalently, z ≠ z₀) is crucial because:

• f may not be defined at z₀, e.g, $f(z) = \frac{\sin z}{z}$ is undefined at z = 0, yet $\lim_{z \to 0}\frac{f(z)}{z} = 1$ exists.
• Limit describes function’s nearby behavior. We want to know what f “wants to be” at z₀.
• Separates limit from value. $\lim_{z \to z_0} f(z)$ may differ from f(z₀). Continuity requires $\lim_{z \to z_0}f(z) = f(z_0)$, but this equality should be a consequence of continuity —not baked into the limit definition itself.
• Enables removable singularities. We can “fill in” a missing point.For $f(z) = \frac{z^2 - 1}{z - 1}$, f is undefined at z = 1, but $\lim_{z \to 1} \frac{z^2 - 1}{z - 1} = \lim_{z \to 1} \frac{(z-1)(z+1)}{z-1} = \lim_{z \to 1} (z + 1) = 2$ We can define $\hat{f}(z) = \begin{cases} f(z), &z \ne 1 \\\\ 2, &z = 1 \end{cases}$.

Remarks

• Uniqueness. If the limit exists, it is unique – no matter what path we take to $z_0$, the limiting value must be the same.
• Path dependence as a test. To show that a limit does not exist, it is often enough to find two different paths along which f(z) approaches different values. For instance, consider $f(z)=\frac{z}{\overline{z}}$ (with $z \neq 0$). Approaching 0 along the real axis gives limit 1, while along the imaginary axis gives limit -1; hence $\lim_{z \to 0}f(z)$ does not exist.
• Relationship with real limits. If we write $z = x+iy$ and $f(z)=u(x,y)+iv(x,y)$, then the complex limit $\lim_{z\to z_0}f(z)=L$ (with L = U + iV) is equivalent to the simultaneous existence of the real limits $\lim_{(x,y)\to(x_0,y_0)} u(x,y)=U,\qquad \lim_{(x,y)\to(x_0,y_0)} v(x,y)=V,$ where the approach is now in the plane. This equivalence is often used to transfer results from real multivariable calculus.

Examples

• A Polynomial Limit. $\lim_{z \to 2+i} z^2 = (2+i)^2 = 3 + 4i$

Let ε > 0. We need to find δ > 0 such that $0 < |z - (2+i)| < \delta \implies |z^2 - (3+4i)| < \varepsilon$

Factor: $|z^2 - (2+i)^2| =[\text{Factor}] |(z - (2+i))(z + (2+i))| = |z - (2+i)| \cdot |z + (2+i)|$

$\delta \le 1$, $|z - (2+i)| < 1$, then by triangle inequality: $|z| \leq |z - (2+i)| + |2+i| < 1 + \sqrt{5}$

Besides, $|z + (2+i)| \leq |z| + |2+i| \lt 1 + \sqrt{5} + \sqrt{5} = 1 + 2\sqrt{5}$

So for $|z - (2+i)| < \delta$ with δ ≤ 1: $|z^2 - (3+4i)| < \delta(1 + 2\sqrt{5})$

Choose $\delta = \min\left(1, \frac{\varepsilon}{1 + 2\sqrt{5}}\right)$. Then, $|z^2 - (3+4i)| < \varepsilon$. ∎

• A piecewise polynomial

Let $f: \overline{B(0; 1)} \rarr \Complex$

f(z) = $\begin{cases} 3z², |z| < 1 \\\\ 3, |z| = 1 \end{cases}$

Continuity at z₀ = 1 and z₀ = -1.

As $z \to 1$ from inside the disk (|z| < 1): $\lim_{z \to 1, |z|<1} f(z) = \lim_{z \to 1} 3z^2 = 3 \cdot 1 = 3$. The value f(1) = 3 matches, so $\lim_{z \to 1} f(z) = f(1) = 3$, so f is continuous at z₀ = 1 and z₀ = -1.

Discontinuity at other boundary points z₀ = eⁱᶿ where θ ≠ 0, π.

From inside the disk: $\lim_{z \to e^{i\theta}, |z|<1} 3z^2 = 3e^{2i\theta}$. However, the boundary value is fixed at f(eⁱᶿ) = 3.

Since $3e^{2i\theta} \neq 3$ when θ ≠ 0, π (since $e^{2i\theta} \neq 1$), the function does not satisfy continuity at these points. Approaching z₀ = eⁱᶿ where θ ≠ 0, π along the boundary (|z| = 1) gives f(z) = 3, while approaching radially from inside gives $3e^{2i\theta}$, so the overall limit $\lim_{z \to e^{i\theta}}f(z)$ does not exist.

• Rational Function with Pole

Let g(z) = $\frac{1}{z-2}, D = \mathbb{C} -${2}.

1. $z \to 2$.
   For any M > 0, choosing $|z - 2| < \frac{1}{M}$, gives $|g(z)| = \frac{1}{|z-2|} > M$.
   The function grows without bound and we say $\lim_{z \to 2} g(z) = \infty$,
   meaning the function diverges to infinity in the extended complex plane (or on the Riemann sphere). No finite limit exists.
2. $z \to \infty$.
   For |z| > 4, we have $|z - 2| \ge |z| - 2 \gt \frac{|z|}{2}$, so: $|g(z)| = \frac{1}{|z-2|} \lt \frac{2}{|z|} \to 0$
   Note: $|z| > 4 \implies |z| - 4 \gt 0 \implies 2|z| -4 \gt |z| \implies |z| - 2 \gt \frac{|z|}{2}$
   Therefore, $\lim_{z \to \infty} g(z) = 0$.

• The Exponential Function For all $z₀ \in \mathbb{C}, \lim_{z \to z_0} e^z = e^{z_0}$

Proof:

1. Let ε > 0. We need: $|e^z - e^{z_0}| < \varepsilon$
2. $|e^z - e^{z_0}| = |e^{z_0}| \cdot |e^{z - z_0} - 1| < \varepsilon$
3. Let $M = |e^{z₀}|$ (a positive constant). We need: $|e^h - 1| < \frac{\varepsilon}{M} \quad \text{where } h = z - z_0$
4. The exponential function is continuous at 0: $\lim_{h \to 0} e^h = 1$.
   We can use the power series of $e^h$ to bound $|e^h - 1|$, $e^h = 1 + h + \frac{h^2}{2!} + \cdots$.
   For |h| < 1: $|e^h - 1| = \left| h + \frac{h^2}{2!} + \frac{h^3}{3!} + \cdots \right| \leq |h| \left(1 + |h| + \frac{|h|^2}{2!} + \cdots \right) \leq |h| \cdot e^{|h|} \leq |h| \cdot e$
   So if $|h| < \min\left(1, \frac{\varepsilon}{eM}\right)$, then $|e^h - 1| < \frac{\varepsilon}{M}$.
5. Choose $\delta = \min\left(1, \frac{\varepsilon}{eM}\right)$. For $0 < |z - z_0| < \delta$: $|e^z - e^{z_0}| = M|e^{z-z_0} - 1| < M \cdot \frac{\varepsilon}{M} = \varepsilon$

• Limit Does Not Exist, $\frac{\bar{z}}{z}$ on D = $\mathbb{C} \setminus \{ 0 \}$

Proof:

1. Write z = reⁱᶿ, so $\bar{z} = re^{-i\theta}$. Then, $f(z) = \frac{re^{-i\theta}}{re^{i\theta}} = e^{-2i\theta}$
2. The value depends only on the argument θ, not on r! In other words, it depends only on direction, not distance.
3. Path 1: Along the positive real axis (θ = 0): $f(z) = e^{0} = 1$
4. Path 2: Along the positive imaginary axis (θ = π/2): $f(z) = e^{-i\pi} = -1$
5. Path 3: Along the line y = x (θ = π/4): $f(z) = e^{-i\pi/2} = -i$
6. Different paths give different values, so no limit exists.

• Branch Cut Discontinuity (-∞, 0]

1. Define $\sqrt{z}$ using the principal branch: $\sqrt{z} = \sqrt{|z|} \, e^{i\arg(z)/2}, \quad \arg(z) \in (-\pi, \pi]$.
2. In order to make $\sqrt{z}$ single-valued, we must choose a branch — and a conventional one is to remove the negative real axis. The branch cut is the ray $(-\infty, 0]$.
3. At points x₀ < 0 on the negative real axis:
4. Approach from above (second quadrant, z = x₀ + iε, $ε \to 0^+$): $\arg(z) \to \pi \implies \sqrt{z} \to \sqrt{|x_0|} \, e^{i\pi/2} = i\sqrt{|x_0|}$
5. Approach from below (third quadrant, z = x₀ - iε, $ε \to 0^-$): $\arg(z) \to -\pi \implies \sqrt{z} \to \sqrt{|x_0|} \, e^{-i\pi/2} = -i\sqrt{|x_0|}$
6. Since $i\sqrt{|x_0|} \neq -i\sqrt{|x_0|}$, the limit does not exist at any point on the branch cut, i.e., a jump across the branch cut. To make multi-valued functions single-valued, we must sacrifice continuity somewhere.

• A Removable Singularity, $f(z) = \frac{\sin z}{z}$ on $\mathbb{C} \setminus \{ 0 \}$

A removable singularity is a point where a function is undefined or not analytic, but can be repaired by redefining its value at that point to make it analytic there. The function $f(z) = \frac{\sin z}{z}$ is undefined at z = 0 due to division by zero.

$\lim_{z \to 0} \frac{\sin z}{z} = 1$

1. Proof using power series. The Taylor series expansion of sin(z) around z = 0 is $\sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots$
2. Dividing by z (for z ≠ 0): $\frac{\sin z}{z} = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots$
3. As $z \to 0$, all terms except the constant 1 vanish, thus $\lim_{z \to 0} \frac{\sin z}{z} = 1$
4. Significance: We can “remove” the singularity at z = 0 by defining f(0) = 1, making f continuous everywhere.
5. Geometric intuition. Imagine a punctured disk around z = 0. The function f(z) behaves nicely as z approaches 0 —it doesn’t blow up or oscillate wildly. The values of f(z) cluster around 1, so defining f(0) = 1 smoothly “plugs the hole” without disrupting the function’s structure.

Limit rules

Suppose D ⊆ ℂ, f, g: D → ℂ, and z₀ is a limit point of D. If $\lim_{z \to z_0} f(z) = L_1$ and $\lim_{z \to z_0} g(z) = L_2$, then the following limit laws hold:

1. Uniqueness of the Limit. If $\lim_{z \to z_0} f(z)$ exists, then it is unique.

   Suppose $\lim_{z \to z_0} f(z) = L$ and $\lim_{z \to z_0} f(z) = M$.
   For any ε > 0, ∃δ₁ > 0: 0 < |z - z₀| < δ₁ ⟹ |f(z) - L| < ε/2, and also ∃δ₂ > 0: 0 < |z - z₀| < δ₂ ⟹ |f(z) - M| < ε/2
   Let δ = min(δ₁, δ₂). Since z₀ is a limit point of D, there exists z ∈ D with 0 < |z - z₀| < δ.
   For such z: $|L - M| = |L - f(z) + f(z) - M| \leq |L - f(z)| + |f(z) - M| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$
   Since ε was arbitrary, |L - M| = 0, so L = M. ∎

2. Linearity. $\lim_{z \to z_0} (c_1 f(z) + c_2 g(z)) = c_1 L_1 + c_2 L_2$

3. Product. $\lim_{z \to z_0} (f(z) · g(z)) = L_1 · L_2$

4. Quotient. $\lim_{z \to z_0} (\frac{f(z)}{g(z)}) = \frac{L_1}{L_2}$ provided L₂ ≠ 0.

5. Conjugate. $\lim_{z \to z_0} \overline{f(z)} = \overline{L_1}$.

6. Modulus. $\lim_{z \to z_0} |f(z)| = |L_1|$

7. Every polynomial p(z) is continuous on $ \mathbb{C}$.
   Proof. By induction on degree, using:
   $\lim_{z \to z_0} c = c$ (constant).
   $\lim_{z \to z_0} z = z_0$ (identity).
   Sum and product rules.

8. Every rational function $r(z) = \frac{p(z)}{q(z)}$ is continuous on its domain {z : q(z) ≠ 0}, e.g., $f(z)=\frac{(3+i)z²+2z-i}{(1+i)z⁴-z³+1}, \lim_{z \to z_0}f(z) = \lim_{z \to z_0}\frac{(3+i)z²+2z-i}{(1+i)z⁴-z³+1} = \frac{(3+i)z_0²+2z_0-i}{(1+i)z_0⁴-z_0³+1} =f(z_0)$

9. The sandwich (squeeze) theorem. If $|f(z) - L| \leq g(z)$ for all z in a punctured neighborhood of z₀ and $\lim_{z \to z_0} g(z) = 0$, then $\lim_{z \to z_0} f(z) = L$ (Note: standard real squeeze theorem requires ordering $f \le g \le h$, which doesn’t exist in $\mathbb{C}$. We squeeze the modulus instead).
   Corollary. If $|f(z)| \leq g(z)$ and $g(z) \to 0$, then $f(z) \to 0$.
   $f(z) = z^2 \sin\left(\frac{1}{|z|}\right)$, we have $|f(z)| \leq |z|^2 \to 0$ as $z \to 0$, so $\lim_{z \to 0} f(z) = 0$.

10. Component-wise Limits. Let f(z) = u(x,y) + iv(x,y) where z = x + iy, and let z₀ = x₀ + iy₀, L = A + iB. Then, $\lim_{z \to z_0} f(z) = L \iff \begin{cases} \lim_{(x,y) \to (x_0, y_0)} u(x,y) = A \\[6pt] \lim_{(x,y) \to (x_0, y_0)} v(x,y) = B \end{cases}$.
    Proof. (⟹) Suppose $\lim_{z \to z_0} f(z) = L$. For any ε > 0, there exists δ > 0 such that: $0 < |z - z_0| < \delta \implies |f(z) - L| < \varepsilon$
    Since: $|u(x,y) - A| = |\text{Re}(f(z) - L)| \leq |f(z) - L| < \varepsilon$, $|v(x,y) - B| = |\text{Im}(f(z) - L)| \leq |f(z) - L| < \varepsilon$, both component limits exist.
    (⟸) Suppose both component limits exist. For ε > 0: ∃δ₁: 0 < |z - z₀| < δ₁ ⟹ |u - A| < ε/√2 and ∃δ₂: 0 < |z - z₀| < δ₂ ⟹ |v - B| < ε/√2
    Let δ = min(δ₁, δ₂). Then, $|f(z) - L| = \sqrt{(u-A)^2 + (v-B)^2} < \sqrt{\frac{\varepsilon^2}{2} + \frac{\varepsilon^2}{2}} = \varepsilon$.
    This theorem reduces complex limits to pairs of real multivariable limits, which can sometimes be easier to evaluate using Calc I-III techniques.

Limits That “Go to Infinity”

Definition. Let $D \subseteq \mathbb{C}, f: D \to \mathbb{C}$, and z₀ be a limit point of D. We say that $\lim_{z \to z_0} f(z) = \infty$ if for every M > 0, there exists δ > 0 such that: $z \in D \text{ and } 0 < |z - z_0| < \delta \implies |f(z)| > M$.

The function f(z) blows up —its modulus values grow without bound as $z \to z₀$. As $z$ gets close to $z_0$, $f(z)$ escapes every bounded disk centered at the origin.

Example. $\lim_{z \to 0} \frac{1}{z} = \infty$

Given M > 0, let δ = 1/M.

For 0 < |z| < $\delta = \frac{1}{M}: \left|\frac{1}{z}\right| = \frac{1}{|z|} > \frac{1}{\delta} = M$

Limits as $z \to \infty$

Definition. Let $D \subseteq \mathbb{C}$ be unbounded, f: $D \to \mathbb{C}$. We say that $\lim_{z \to \infty} f(z) = L$ if for every ε > 0, there exists M > 0 such that: $z \in D \text{ and } |z| > M \implies |f(z) - L| < \varepsilon$

For z with large modulus, f(z) is close to L. In other words, as $z$ moves far away from the origin in any direction, $f(z)$ settles close to $L$.

Example: $\lim_{z \to \infty} \frac{1}{z} = 0$

Proof. Given ε > 0, let M = 1/ε.

For |z| > M: $\left|\frac{1}{z} - 0\right| = \frac{1}{|z|} < \frac{1}{M} = \varepsilon$

• Rational Function Limit at Infinity. $\lim_{z \to \infty} \frac{3z²}{(1+i)z²-z+2} = \frac{3}{1+i} = \frac{3(1-i)}{2}$

  Both numerator and denominator are degree-2 polynomials. The limit simplifies to the ratio of leading coefficients.

$|\frac{3z²}{(1+i)z²-z+2} - \frac{3}{1+i}| = |\frac{3z²(1+i)-3z²(1+i)+3z-6}{(1+i)((1+i)z²-z+2)}| = |\frac{3z-6}{(1+i)((1+i)z²-z+2)}| =[\text{Then, dividing both numerator and denominator by z² within the absolute value}]$

$|\frac{\frac{3}{z}-\frac{6}{z²}}{(1+i)((1+i)-\frac{1}{z}+\frac{2}{z²})}|$ [*]

$|(1+i)-\frac{1}{z}+\frac{2}{z²}| \ge[\text{Recall the reverse triangle inequality}|a-b| \ge |a| - |b|]\sqrt{2}-|\frac{1}{z}-\frac{2}{z²}|$

$|z| \ge M \leadsto \frac{1}{|z|} \le \frac{1}{M} \leadsto [\text{Triangular inequality}] |\frac{1}{z}-\frac{2}{z²}| \le |\frac{1}{z}| + |\frac{-2}{z^2}| = |\frac{1}{z}| + |\frac{2}{z^2}| \le \frac{1}{M}+\frac{2}{M²}$

$|(1+i)-\frac{1}{z}+\frac{2}{z²}| \ge \sqrt{2}-|\frac{1}{z}-\frac{2}{z²}| \ge \sqrt{2}-(\frac{1}{M}+\frac{2}{M²})$.

Bounding the Entire Expression: [*] $\le \frac{\frac{3}{M}+\frac{6}{M²}}{\sqrt{2}(\sqrt{2}-\frac{1}{M}-\frac{2}{M²})}$

For large M, approximate: (i) Numerator: $\frac{3}{M}+\frac{6}{M²} \le \frac{3}{M} + \frac{6}{M} =\frac{9}{M}$; (ii) $\sqrt{2}(\sqrt{2}-\frac{1}{M}-\frac{2}{M²}) \ge \sqrt{2}\cdot \sqrt{2} = 2$

Refining the previous bond: $\le \frac{\frac{9}{M}}{2} = \frac{9}{2M} \le \varepsilon$, e.g., $M \ge \frac{9}{2\varepsilon}$

For any $\varepsilon \gt 0$, choose $M \ge \frac{9}{2\varepsilon}$, then |z| ≥ $\frac{9}{2\varepsilon}, |\frac{3z²}{(1+i)z²-z+2} - \frac{3}{1+i}| \le \frac{9}{2M} \le \varepsilon$.

Definition. $\lim_{z \to \infty} f(z) = \infty$ if $\forall N > 0, \; \exists M > 0 : |z| > M \implies |f(z)| > N$

Example: $\lim_{z \to \infty} z^2 = \infty$

For |z| > M = $\sqrt{N}$, we have |z²| = |z|² > N.

The Riemann Sphere Perspective

The extended complex plane $\hat{\mathbb{C}} = \mathbb{C} \cup \{\infty\}$ can be visualized as a sphere (the Riemann sphere) via stereographic projection.

On this sphere (∞) is a genuine point (the “north pole”), limits as $z \to \infty$ are ordinary limits toward this point. Therefore, the function f(z) = 1/z extends to a continuous map on the sphere with f(0) = ∞ and f(∞) = 0.

Complex Rational Limits at Infinity

When dealing with rational functions $R(z) = \frac{P(z)}{Q(z)}$, the behavior at infinity is determined by the degrees of the polynomials.

Let $P(z) = a_n z^n + \dots$ and $Q(z) = b_m z^m + \dots$ with $a_n, b_m \neq 0$.

$$ \lim_{z \to \infty} \frac{P(z)}{Q(z)} = \begin{cases} \infty & \text{if } n > m \\ \frac{a_n}{b_m} & \text{if } n = m \\ 0 & \text{if } n < m \end{cases} $$