Calculus/Infinite Limits/Infinity is not a number

Most people seem to struggle with this fact when first introduced to calculus, and in particular limits.

${\displaystyle \lim _{x\to 0^{+}}{\frac {1}{x}}=\infty }$

But ${\displaystyle \infty }$ is different. ${\displaystyle \infty }$ is not a number.

Mathematics is based on formal rules that govern the subject. When a list of formal rules applies to a type of object (e.g., "a number") those rules must always apply — no exceptions!

What makes ${\displaystyle \infty }$ different is this: "there is no number greater than infinity". You can write down the formula in a lot of different ways, but here's one way: ${\displaystyle 1+\infty =\infty }$ . If you add one to infinity, you still have infinity; you don't have a bigger number. If you believe that, then infinity is not a number.

Since ${\displaystyle \infty }$ does not follow the rules laid down for numbers, it cannot be a number. Every time you use the symbol ${\displaystyle \infty }$ in a formula where you would normally use a number, you have to interpret the formula differently. Let's look at how ${\displaystyle \infty }$ does not follow the rules that every actual number does:

Every number has a negative, and addition is associative. For ${\displaystyle \infty }$ we could write ${\displaystyle -\infty }$ and note that ${\displaystyle \infty -\infty =0}$ . This is a good thing, since it means we can prove if you take one away from infinity, you still have infinity: ${\displaystyle \infty -1=(\infty +1)-1=\infty +(1-1)=\infty +0=\infty }$ . But it also means we can prove 1 = 0, which is not so good.

{\displaystyle {\begin{aligned}&1+\infty =\infty \\&(1+\infty )-\infty =\infty -\infty \\&1+(\infty -\infty )=\infty -\infty \\&1=0\end{aligned}}}

Therefore, ${\displaystyle \infty -\infty ={\text{indeterminate}}}$ .

Reinterpret Formulae that Use ${\displaystyle \infty }$

We started off with a formula that does "mean" something, even though it used ${\displaystyle \infty }$ and ${\displaystyle \infty }$ is not a number.

${\displaystyle \lim _{x\to 0^{+}}{\frac {1}{x}}=\infty }$

What does this mean, compared to what it means when we have a regular number instead of an infinity symbol:

${\displaystyle \lim _{x\to 2}{\frac {1}{x}}={\frac {1}{2}}}$

This formula says that I can make sure the values of ${\displaystyle {\frac {1}{x}}}$ don't differ very much from ${\displaystyle {\tfrac {1}{2}}}$ , so long as I can control how much ${\displaystyle x}$ varies away from 2. I don't have to make ${\displaystyle {\frac {1}{x}}}$ exactly equal to ${\displaystyle {\tfrac {1}{2}}}$ , but I also can't control ${\displaystyle x}$ too tightly. I have to give you a range to vary ${\displaystyle x}$ within. It's just going to be very, very small (probably) if you want to make ${\displaystyle {\frac {1}{x}}}$ very very close to ${\displaystyle {\tfrac {1}{2}}}$ . And by the way, it doesn't matter at all what happens when ${\displaystyle x=2}$ .

If we could use the same paragraph as a template for my original formula, we'll see some problems. Let's substitute 0 for 2, and ${\displaystyle \infty }$ for ${\displaystyle {\frac {1}{2}}}$ .

${\displaystyle \lim _{x\to 0^{+}}{\frac {1}{x}}=\infty }$

This formula says that I can make sure the values of ${\displaystyle {\frac {1}{x}}}$ don't differ very much from ${\displaystyle \infty }$ , so long as I can control how much ${\displaystyle x}$ varies away from 0. I don't have to make ${\displaystyle {\frac {1}{x}}}$ exactly equal to ${\displaystyle \infty }$ , but I also can't control ${\displaystyle x}$ too tightly. I have to give you a range to vary ${\displaystyle x}$ within. It's just going to be very, very small (probably) if you want to see that ${\displaystyle {\frac {1}{x}}}$ gets very, very close to ${\displaystyle \infty }$ . And by the way, it doesn't matter at all what happens when ${\displaystyle x=0}$ .

It's close to making sense, but it isn't quite there. It doesn't make sense to say that some real number is really "close" to ${\displaystyle \infty }$ . For example, when ${\displaystyle x=0.001}$ and ${\displaystyle {\frac {1}{x}}=1000}$ does it really makes sense to say 1000 is closer to ${\displaystyle \infty }$ than 1 is? Solve the following equations for ${\displaystyle \delta }$ :

{\displaystyle {\begin{aligned}&1000+\delta =\infty \\&\delta =\infty -1000\\&\delta =\infty \end{aligned}}}
{\displaystyle {\begin{aligned}&1+\delta =\infty \\&\delta =\infty -1\\&\delta =\infty \end{aligned}}}

No real number is very close to ${\displaystyle \infty }$; that's what makes ${\displaystyle \infty }$ so special! So we have to rephrase the paragraph:

${\displaystyle \lim _{x\to 0^{+}}{\frac {1}{x}}=\infty }$

This formula says that I can make sure the values of ${\displaystyle {\frac {1}{x}}}$ get as big as any number you pick, so long as I can control how much ${\displaystyle x}$ varies away from 0. I don't have to make ${\displaystyle {\frac {1}{x}}}$ bigger than every number, but I also can't control ${\displaystyle x}$ too tightly. I have to give you a range to vary ${\displaystyle x}$ within. It's just going to be very, very small (probably) if you want to see that ${\displaystyle {\frac {1}{x}}}$ gets very, very large. And by the way, it doesn't matter at all what happens when ${\displaystyle x=0}$ .

You can see that the essential nature of the formula hasn't changed, but the exact details require some human interpretation. While rigorous definitions and clear distinctions are essential to the study of mathematics, sometimes a bit of casual rewording is okay. You just have to make sure you understand what a formula really means so you can draw conclusions correctly.

Exercises

Write out an explanatory paragraph for the following limits that include ${\displaystyle \infty }$ . Remember that you will have to change any comparison of magnitude between a real number and ${\displaystyle \infty }$ to a different phrase. In the second case, you will have to work out for yourself what the formula means.

1. ${\displaystyle \lim _{x\to \infty }{\frac {1}{x^{2}}}=0}$

This formula says that I can make the values of ${\displaystyle {\frac {1}{x^{2}}}}$ as close as I would like to 0, so long as I make ${\displaystyle x}$ sufficiently large.

2. ${\displaystyle \sum _{n=0}^{\infty }2^{-n}=1+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+\cdots =2}$

This formula says that you can make the sum ${\displaystyle \sum _{k=0}^{n}2^{-k}}$ as close as you would like to 2 by making ${\displaystyle n}$ sufficiently large.