I'm a fifteen year old currently studying calculus and our teacher today explained the properties of infinite limits. Firstly, these were intriguing, since they appeared to imply that one can do algebraic manipulation with infinity, something I've long been told cannot be done, because "infinity is a concept, not a number." One of the properties was particularly interesting:

If lim(f(x)) = infinity, and lim(g(x)) = L, then the following is true: x->c x->c

Lim(g(x)/f(x)) = 0 x->c

Now, this basically is saying that the limit of 1/infinity = 0, which makes sense and works with both positive and negative infinity (1/-1000000000 is very close to 0, therefore the limit of 1/-infinity should also be equal to 0). Therefore, I introduce a new function h(x), whose limit is thus:

Lim(h(x)) = -infinity x->c

Using the property mentioned,

Lim(g(x)/h(x)) = 0 x->c

Now, since both limits are equal to 0, they are equal to one another. (a=b, b=c, thus a=c)

Lim(g(x)/h(x)) = Lim(g(x)/f(x)) x->c x->c

Using the quotient property of limits (Lim(x/y) = lim(x)/lim(y)), we get this:

Lim(g(x))/Lim(h(x)) = Lim(g(x))/Lim(f(x)) x->c x->c x->c x->c

Cross-multiplying yields the following result:

Lim(g(x)) * Lim(f(x)) = Lim(g(x)) * Lim(h(x)) x->c x->c x->c x->c

Using the original definitions of these limits, in which the limit of f(x) at x-> c is infinity, h(x) at x->c is -infinity, and the g(x) at x->c is L, we get:

L * infinity = L * -infinity

And since anything times infinity is just infinity, we get:

infinity = -infinity

This startling result was very interesting. I don't think I made any mathematical mistake, so the only two things that I can think of that fix this issue are the following:

  1. The limits of 1/infinity and 1/-infinity are not both 0, rather the 1/infinity is +0 and 1/-infinity is -0.


  1. The limits are correct and infinity = -infinity, in which case the number line is a circle and infinity and -infinity are the same.

Help and critique on my logic is requested, as well as an explanation from a more experienced mathematician.

  • 1
    $\begingroup$ You did make mathematical mistakes. You cannot treat infinity just like any other number. Addition and multiplication are defined on the real numbers and infinity is not a real number. Also infinity in limits is actually shorthand and the formal definition doesn't reference infinity at all. $\endgroup$ – lordoftheshadows Feb 15 '17 at 1:07

There's a subtle mathematical error present - the limit law $\lim_{x\to c}\frac{f(x)}{g(x)} = \frac{\lim_{x\to c}f(x)}{\lim_{x\to c}g(x)}$ requires that both limits exist and that $\lim_{x\to c}g(x)$ not be zero. But a limit that goes to infinity doesn't count as "existing", so the limit law doesn't apply.

The other issue is that you were right at the beginning - you can't manipulate infinity algebraically. Your result at the end is sort of a proof of that - it results in something that doesn't seem right based on algebraically manipulating infinity. In particular, cross-multiplying doesn't work when there's an infinity in the mix.

The purpose of a limit is to allow you to treat $\infty$ as a number as much as that is possible. But you have to use limits for everything you want to do with infinity - as soon as you strip away the $\lim$, you're back to using it as an actual number, which can't work. For example, to cross-multiply two limits of fractions, I have to do this:

$\lim_{x \to c}\frac{f(x)}{g(x)} = \lim_{x \to c}\frac{h(x)}{k(x)}$

$\frac{\lim_{x \to c}f(x)}{\lim_{x \to c}g(x)} = \frac{\lim_{x \to c}h(x)}{\lim_{x \to c}k(x)}$

$\lim_{x \to c}f(x)\lim_{x \to c}k(x) = \lim_{x \to c}h(x)\lim_{x \to c}g(x)$

$\lim_{x \to c}f(x)k(x) = \lim_{x \to c}h(x)g(x)$

This required me to use the division law and the multiplication law, each twice. Each one of those requires that all of the limits exist - so none of this works if any of $\lim_{x \to c}f(x)$, $\lim_{x \to c}g(x)$, $\lim_{x \to c}h(x)$, or $\lim_{x \to c}k(x)$ are $\pm\infty$.

Keep an eye out for situations like this, though - spotting things like this can help you see why the obscure rules in the limit laws are important.

  • $\begingroup$ Well explained (+1) $\endgroup$ – Juniven Feb 15 '17 at 1:14
  • $\begingroup$ Nice explanation. I saw this a lot when I was a TA for a high school calculus class and when helping friends. I wonder if it would be a good idea to emphasize the concepts hiding behind our notational shorthand of infinity. Maybe introduce epsilon delta in a first calculus course so as not to build the wrong kind of intuition. $\endgroup$ – lordoftheshadows Feb 15 '17 at 1:18

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