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The real definition of a limit with the modulus in place of the absolute value, and the consequence that matters: a limit must agree along every approach, so two directions can disprove one and can never prove one.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state the definition of a limit and of continuity for a complex function, evaluate a limit by continuity or by cancelling a common factor, show that a limit does not exist by finding two approaches that disagree, say why two approaches that agree prove nothing, and split a complex limit into two real ones.
Limits of functions of one real variable, and the modulus as a distance in the plane. The definition below is the real one with distance on the line replaced by distance in the plane, and that single substitution is what makes the subject rigid: there are infinitely many directions of approach instead of two.
A neighbourhood of $z_0$ is an open disc about it; a punctured neighbourhood leaves $z_0$ out. $f$ has limit $L$ at $z_0$ when the values $f(z)$ can be forced within any distance of $L$ by taking $z$ close enough to $z_0$ but not equal to it. $f$ is continuous at $z_0$ when that limit exists and equals $f(z_0)$. A domain is an open connected set, and it is the setting every theorem in this course is stated on.
The definition. $\lim_{z \to z_0} f(z) = L$ means: for every $\varepsilon > 0$ there is a $\delta > 0$ such that
$$0 < |z - z_0| < \delta \implies |f(z) - L| < \varepsilon.$$
Word for word the real definition, with $|\cdot|$ now the modulus. Everything algebraic carries over unchanged: limits of sums, products and quotients behave as they did, polynomials are continuous everywhere, and a rational function is continuous wherever its denominator is not zero.
What does not carry over is how much the definition asks for. On the line, $z \to z_0$ has two sides. In the plane, $|z - z_0| < \delta$ is a disc, and the condition has to hold for every point of it — along every ray, every spiral, every wandering path. A limit in the plane is a statement about infinitely many approaches at once.
This cuts both ways, and the asymmetry is the thing to remember:
The standard example. $f(z) = \bar z / z$ for $z \ne 0$. Along the real axis $\bar z = z$, so $f = 1$. Along the imaginary axis $\bar z = -z$, so $f = -1$. No limit at the origin, and no value at the origin can make $f$ continuous.
Splitting into parts. Writing $f = u + iv$ and $L = A + iB$, the limit exists and equals $L$ exactly when $u \to A$ and $v \to B$ as real functions of two real variables. So a complex limit is two real limits, and either may be attacked with whatever is known about functions of two variables.
Another way: picture
Draw a small disc about $z_0$ and ask what $f$ does to it. A limit exists when those images shrink to a point as the disc shrinks — not when they shrink along two chosen lines. For $\bar z / z$ the image of every disc about the origin, however small, is the whole unit circle: shrinking the disc does not shrink the image at all, which is as far from having a limit as a function can get.
Another way: steps
It is tempting to treat the limit along the real axis and the limit along the imaginary axis as the complex version of a left-hand and a right-hand limit, and to conclude that if they agree the limit exists. They are not, and it does not.
Take $f(z) = \dfrac{(\operatorname{Re} z)(\operatorname{Im} z)}{|z|^{2}}$ away from the origin. On the real axis the numerator is zero, so $f = 0$; on the imaginary axis the numerator is zero again, so $f = 0$. Two directions, perfect agreement. But along the line $y = x$, the numerator is $x^{2}$ and the denominator $2x^{2}$, so $f = \tfrac12$ the whole way in. There is no limit.
On the line, two one-sided limits exhaust the possibilities and so their agreement is a theorem. In the plane they exhaust nothing. The habit worth building is to treat two agreeing directions as a reason to look for a proof, never as the proof — and the proof, when it comes, is almost always a bound on $|f(z) - L|$ in terms of $|z - z_0|$ alone, because such a bound cannot tell the directions apart.
This is also the first hint of why the next lesson's definition is so strong. A derivative is a limit, and a limit in the plane has to survive every approach. Asking that of a difference quotient turns out to be an enormous demand.
The single most common error here is to compute along the real axis, compute along the imaginary axis, find the same answer, and write therefore the limit is. The example in the section above shows why that is not an argument: a function can agree on both axes and take a completely different constant value along the diagonal between them.
The reverse error is rarer and also worth naming: having found two directions that disagree, some people go on to check a third. There is no need. One disagreement settles it, because a limit is required to be the same along every approach and these two already are not.
A third habit to break is treating $|z| \to 0$ and $z \to 0$ as different statements. They are the same statement: $|z - z_0|$ is the only measure of closeness the plane has, and every limit here is ultimately a statement about that one real number becoming small.
$\lim_{z \to 2i} \dfrac{z^{2} + 4}{z - 2i}$: the top is $(z - 2i)(z + 2i)$.
Factorise before panicking.
Away from $2i$ the quotient is $z + 2i$, which is continuous, so the limit is $2i + 2i = 4i$.
Cancel, then substitute.
$\lim_{z \to 0} \dfrac{\operatorname{Re} z}{z}$: along the real axis, $\operatorname{Re} z = z$ and the quotient is $1$.
First approach.
Along the imaginary axis, $\operatorname{Re} z = 0$ and the quotient is $0$.
Second approach.
The two disagree, so no limit exists. Note that one disagreement was enough — no further directions need checking.
Disproving needs two; proving needs all of them.
$|z|^{2} = z\bar z$, so away from the origin the quotient is $\bar z$.
Rewrite before testing.
$\bar z \to 0$ as $z \to 0$, along every approach, so the limit exists and is $0$.
Let $f(z) = z^{2} + z$. Find the limit of $f$ as $z$ tends to $6 + 3i$ by filling in the table.
| Value | |
|---|---|
| The real part of the square of the point | |
| The imaginary part of the square of the point | |
| The real part of the limit | |
| The imaginary part of the limit |
$g(z) = \dfrac{z^{2} - 9}{z - 3}$ is undefined at $z = 3$. What value there makes it continuous?
Answer:
Select every statement that is true.
This task has no paper form; do it on a device.
Each function has a hole. Match it to the value that fills the hole.
| $18$ | $0$ | $9$ | $-18$ | |
|---|---|---|---|---|
| $\dfrac{z^{2} - 81}{z - 9}$ at $z = 9$ | ||||
| $\dfrac{z^{3}}{z^{2}}$ at $z = 0$ | ||||
| $\dfrac{z^{2} + 9z}{z}$ at $z = 0$ | ||||
| $\dfrac{z^{2} - 81}{z + 9}$ at $z = -9$ |
Put the steps of showing that $\dfrac{\bar z}{z}$ has no limit at the origin in order.
Number the steps in order (write the number in the box):
What does $\dfrac{3z^{2} + 8z + 1}{5z^{2} + 1}$ tend to as $|z|$ grows without bound? Give a fraction.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Let $f(z) = z^{2} + 6z$. Find the limit of $f$ as $z$ tends to $4 + 5i$ by filling in the table.
| Value | |
|---|---|
| The real part of the square of the point | |
| The imaginary part of the square of the point | |
| The real part of the limit | |
| The imaginary part of the limit |
You can evaluate limits where they exist and disprove them where they do not, and you can say what the plane demands that the line did not. Say in your own words why the conjugate divided by the number has no limit at the origin. Next: the derivative, which is a limit of exactly this kind and is therefore a far stronger condition than it looks.
9. Your turn: does $\dfrac{|z|^{2}}{z}$ have a limit at the origin?, step 2