Wolfram Alpha says: $$i\lim_{x \to \infty} x = i\infty$$

I'm having a bit of trouble understanding what $i\infty$ means. In the long run, it seems that whatever gets multiplied by $\infty$ doesn't really matter. $\infty$ sort of takes over, and the magnitude of whatever is being multiplied is irrelevant. I.e., $\forall a \gt 0$:

$$a\lim_{x \to \infty} x = \infty, -a\lim_{x \to \infty} x = -\infty$$

What's so special about imaginary numbers? Why doesn't $\infty$ take over when it gets multiplied by $i$? Thanks.

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    $\begingroup$ I think by $a\ne 0$ you meant $a>0$, otherwise the signs on your $\pm\infty$ could be wrong. $\endgroup$
    – Mark S.
    May 14, 2017 at 3:32
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    $\begingroup$ is $x$ real here? $\endgroup$
    – pjpj
    May 14, 2017 at 3:37
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    $\begingroup$ Always remember, Wolfram Alpha is merely an approximation of mathematics. $\endgroup$ May 14, 2017 at 3:39
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    $\begingroup$ Think of the complex plane. (Also, $i$ is neither positive nor negative.) $\endgroup$ May 14, 2017 at 5:18
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    $\begingroup$ You could also ask: what's so special about $-1$ that $-1\cdot \lim_{x\to\infty} x = -1\cdot \infty$? $\endgroup$
    – M. Winter
    Feb 2, 2020 at 17:42

5 Answers 5


In the reals all non-zero numbers have a parity. Either they are positive or they are negative. $\lim_{x\rightarrow \infty}|ax| =\infty$ (if $a \ne 0$) because the magnitude of $ax$ gets infinitely large.

If $a > 0$ then $\lim_{x\rightarrow \infty}ax = \infty$ because the magnitude of $ax$ becomes infinite and the parity of all $ax$ is positive.

If $-a < 0$ then $\lim_{x\rightarrow \infty}-ax = -\infty$. What's the difference between $-\infty$ and $\infty$? Neither of them are actual numbers. Well, again, the magnitude of $-ax$ becomes infinite. But the parity of all $-ax$ is negative so instead of increasing infinitely "in the positive direction", $-ax$ increase in the "negative direction". So $-\infty $ indicates infinite magnitude- negative parity.

Non-zero Complex numbers do not have a single bidirectional parity. A complex number has two components, a real one and an imaginary on and thus are two-dimensional and instead of having a single positive/negative parity, they have a directional angle called an argument. These angles can be in any of an infinite number of "directions" from $0^{\circ}$ to $360^{\circ}$. The number positive $1$ has an argument of $0^{\circ}$. Then number $-1$ has an argument of $180^{\circ}$. The number $\frac 12 + i \frac {\sqrt{3}}2$ has an argument of $30^{\circ}$.

And $i$ has argument $90^{\circ}$.

So what happens to $ix$ as $x \rightarrow \infty$? Well just line $ax$ and $-ax$ its magnitude increases to infinity. So $\lim_{x\to\infty} |ix| = \infty$. But what is the argument of all the $ix$? They all have an argument of $90^{\circ}$. But $\infty$ means infinite magnitude-positive parity. And $-\infty$ means infinite magnitude- negative parity.

Neither of those apply for $\lim_{x\to\infty} ix$ which will have infinite magnitude - $90^{\circ}$ argument. How can we indicate that?

Well.... if $-\infty$ means negative parity and $+\infty$ means positive parity, shouldn't $i\infty$ mean $90^{\circ}$ argument?

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    $\begingroup$ "-ax increase in the negative direction" should this be decrease then? $\endgroup$
    – giraffesyo
    May 14, 2017 at 14:44
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    $\begingroup$ "should this be decrease then?" The real numbers in the ordered field are decreasing, yes, but that is irrelevant. I'm talking about the magnitudes (absolute values) of the numbers. And the magnitudes are increasing.... with direction. The order of the real numbers only exists for the real numbers. The complex numbers have no order and order is only one dimensional. $\endgroup$
    – fleablood
    May 14, 2017 at 16:00
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    $\begingroup$ In your third sentence, you write that the limit tends to infinity; that's not right (according to most authors). $\endgroup$ May 14, 2017 at 18:59
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    $\begingroup$ I disagree with your use of the term "parity", which stands for class modulo$~2$ of integers (related to grouping into pairs). Maybe one of "direction", "sign" or such would do better? $\endgroup$ May 15, 2017 at 9:48
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    $\begingroup$ "I disagree with your use of the term "parity", which stands for class modulo 2 of integers" Oops. You are correct. But then again "parity" in english only means any class that has two distinct and exhaustive states. $\endgroup$
    – fleablood
    May 15, 2017 at 16:20

Briefly, Wolfram|Alpha preserves the $i$ because it's giving you a "direction" for the infinity. Just like $-7\displaystyle{\lim_{x\to\infty}}x=-\infty$ (the direction being "leftwards" on the real line/in the complex plane), $i\displaystyle{\lim_{x\to\infty}}x=i\infty$ (the direction being "upwards" in the complex plane).

This can be used to do many things. To steal an example from the documentation for DirectedInfinity, you can ask Wolfram|Alpha to find an approximation of $\arcsin$ for numbers of the form $\varepsilon+yi$ where $|\varepsilon|$ is very small and $y$ is a very large positive number. Wolfram|Alpha link

This is contrasted with Wolfram|Alpha's confusingly-named "complex infinity", where there is no particular direction. See ComplexInfinity for documentation.

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    $\begingroup$ Complex infinity comes from the Riemann sphere; it is the singular point at infinity, or the antipode of zero. $\endgroup$
    – Kevin
    May 15, 2017 at 3:27

Real numbers and imaginary numbers are different things. is different from ∞i, just like infinity oranges and infinity bottles of juice are different things. There are operations that will convert one to the other, but that is beside the point.

I think a good way to see this is to see where the reals and imaginaries lie on the Argand Diagram (complex numbers and nonintegers omitted for brevity / clarity.)

enter image description here

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    $\begingroup$ Great analogy on differentiation of infinities lol. Very useful graph diagram also! $\endgroup$
    – DukeZhou
    Sep 19, 2017 at 2:38
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    $\begingroup$ Diagram doesn't look right on mobile. Just in case you'd like to edit later $\endgroup$
    – Yuriy S
    Feb 1, 2018 at 0:57

Perhaps this broader view can help. (Perhaps it confuses instead!) There are several different ways to describe ways to go to infinity. I would identify these ways with different compactifications. Compactification is a formalized way of adding points at infinity.

Let us first study the real line. The most common compactification is the two-point compactification, in which we add $+\infty$ and $-\infty$. This means that we make a difference between infinities at the two directions, but only add one "limit point" at each end. Another alternative is to only add one infinity (call it $\tilde\infty$), and say that $x_i\to\tilde\infty$ whenever $|x_i|\to\infty$ (in the usual sense).

You can also add several infinities in both directions. The maximal compactification (the most infinities you can consistently add) is the Stone–Čech compactification which is horribly large: there are more infinities than real numbers. The infinity can somehow branch in a peculiar way, but I will not go any deeper here. This is just to show that you can consider far more exotic infinities if you want to.

Let us then turn to the complex plane. The most common compactification is the one-point one (known as the Riemann sphere), where a single infinity $\tilde\infty$ is added. In this compactification $ki$ tends to $\tilde\infty$ as $k\to\infty$.

One alternative is a radial compactification, adding one point at each direction. (Formally, you can take a radial diffeomorphism of the complex plane to the open unit disc, and the compactification will be the closed disc.) In this compactification you can describe infinities as $\lambda\infty$, where $\lambda$ is a complex number of unit length. In this compactification $ki$ tends to $i\infty$ as $k\to\infty$.

The complex plane also has Stone–Čech compactification. Something yet different happens to your sequence there. (The space $\beta\mathbb C$ is not sequentially compact unless I'm mistaken, and I'm not sure if this particular sequence even has a limit. But that's not even important for this answer.)

So, the limit $\lim_{k\to\infty}ik$ can be different things depending on how you view the infinity. There is no single correct answer. The same applies to $i\lim_{k\to\infty}k$, although multiplication does not always extend nicely to limit objects. In the two reasonable compactifications of the complex plane I presented, multiplication makes sense and commutes with the limit. For the radial compactification $i\lim_{k\to\infty}k$ is indeed $i\infty$, but for the one-point one it is $i\tilde\infty=\tilde\infty$.


In Mathematica, evaluating

Limit[x, x -> Infinity]

gives (the usual shorthand symbol for the build-in entity) Infinity. No problem there. And also in Mathematica, evaluating

I Infinity

must returns as output the same thing you entered (albeit with the shorthand symbol for Infinity and the stylized i representing that complex number).

What this means is that no further evaluation is possible: Mathematica knows no further rules that would allow it to simplify the result further.

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    $\begingroup$ You say, "evaluating I Infinity must return as output the same thing you entered". But why? It certainly doesn't do that for 2 * Infinity. You say, "Mathematica knows no further rules that would allow it to simplify the result further.". But again: why? Those rules were chosen somehow, presumably by a person; why didn't that person create a rule for simplifying this? Just waving your hands and saying "well Mathematica doesn't know any better" doesn't really answer anything. $\endgroup$ May 14, 2017 at 22:26
  • $\begingroup$ The Mathematica reason: because Mathematica doesn't know any rule for how to simplify I Infinity. The mathematical reason: because there's no good way to decide what the product of the complex unit i and the symbol for infinity should mean (and further, because if you want to consider the product to be the limit of I * x as x tends to infinity, there's no preferred direction in which to let x move). $\endgroup$
    – murray
    May 16, 2017 at 0:37

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