Limits of complex functions work like limits in two variables: $z$ can approach $z_0$ from any direction and along any curve, not just from the left and the right. That freedom is what makes the complex derivative so demanding, so it is worth seeing clearly first.
Idea clave
$\displaystyle\lim_{z \to z_0} f(z) = L$ means $|f(z) - L|$ can be made as small as we like by taking $0 < |z - z_0|$ small enough, whatever the direction. So if two paths to $z_0$ give two different values, the limit does not exist. $f$ is continuous at $z_0$ when $\lim_{z \to z_0} f(z) = f(z_0)$. Sums, products, quotients (with a nonzero bottom) and compositions of continuous functions are continuous, so polynomials are continuous everywhere.
Por qué funciona
The definition only mentions the distance $|z - z_0|$, which is the same in every direction, so it cannot prefer one path. In terms of real and imaginary parts, $f(z) \to L$ exactly when $u \to \operatorname{Re} L$ and $v \to \operatorname{Im} L$, because $|f - L|^2 = (u - \operatorname{Re} L)^2 + (v - \operatorname{Im} L)^2$. To show a limit exists, bound $|f(z) - L|$ by something that tends to $0$; to show it does not, find two paths that disagree.
Worked examples
Ejemplo resuelto: A polynomial
Find $\displaystyle\lim_{z \to 1 + i}(z^2 - 3z)$.
- Polynomials are continuous, so substitute.
- $(1 + i)^2 - 3(1 + i) = 2i - 3 - 3i = -3 - i$.
Respuesta: $-3 - i$
Ejemplo resuelto: Zero over zero
Find $\displaystyle\lim_{z \to 2i} \frac{z^2 + 4}{z - 2i}$.
- Substituting gives $\frac{0}{0}$, so $z - 2i$ is a factor of the top: $z^2 + 4 = (z - 2i)(z + 2i)$.
- For $z \ne 2i$ the quotient is $z + 2i$, which tends to $4i$.
Respuesta: $4i$
Ejemplo resuelto: A limit that does not exist
Does $\displaystyle\lim_{z \to 0} \frac{\overline{z}}{z}$ exist?
- Along the real axis, $z = x$: $\frac{x}{x} = 1$.
- Along the imaginary axis, $z = iy$: $\frac{-iy}{iy} = -1$.
- Two paths, two values: the limit does not exist. (Along the line $y = x$ the value is $\frac{1 - i}{1 + i} = -i$, a third value.)
Respuesta: no
Ejemplo resuelto: A limit that does exist
Find $\displaystyle\lim_{z \to 0} \frac{\overline{z}^{\,2}}{z}$.
- Bound the modulus: $\left|\frac{\overline{z}^{\,2}}{z}\right| = \frac{|z|^2}{|z|} = |z|$.
- $|z| \to 0$ along every path, so the limit is $0$.
Respuesta: $0$
Error común
Checking only the real axis, or only straight lines. Agreement along the real axis proves nothing; agreement along every line through $z_0$ still is not enough in general (as in calculus 3, a curve such as a parabola can disagree with all lines). To prove a limit exists, bound $|f(z) - L|$ by a quantity that tends to $0$, such as $|z - z_0|$. To disprove it, two paths with different values are enough.