# Why is $\frac{\ln\infty}{\infty}$ equal to $\frac\infty\infty$?

For me there is no intuitive explanation for this. Yes, i get that if you want to find the natural logarithm of a very high number (infinity) that the ln would be high too. But it does not grow nearly as fast as infinity itself, for example the natural logarithm of $\ln{100} \approx 4.60$ that number is 21 times as small as the number we are taking ln off. But anyhow, here comes the equation making my head jittery.

$$\lim_{x\to\infty} \frac{\ln{\left(x\right)}}{x}=\frac{\infty}{\infty}$$

How, is this possible. Doesn't $\ln$ infinity grow at a much slower speed than infinity itself?

I know how to do derivatives. I'm asking for an intuitive explanation. Of why we use that specific convention.

In other words, why do we use an indeterminate form, and is there an intuitive explanation for why we might use infinity over infinity, or zero over zero?

Khans example. Showing that $\frac{ln\left(x\right)}{x} = \frac{\infty}{\infty}$

• It's not. This limit tends to $0$. You are thinking correctly. – Kaynex Nov 16 '16 at 19:41
• Infinity is not "a very high number"; it is not a number at all, in the context of calculus. Doing arithmetic with it, like $\frac{\infty}{\infty}$ is a mistake; the correct way to handle infinity is via limits (or other advanced methods from logic). – vadim123 Nov 16 '16 at 19:41
• This is a very confused question... – DonAntonio Nov 16 '16 at 19:42
• That Khan academy image suggests one way to make sense of the ratio of the two infinities, but not the best way. Your intuition is better, as is @AlexM 's answer. – Ethan Bolker Nov 16 '16 at 19:51
• @AlexM. Derivatives is not the problem per se, i know how how to do them. And i've completed the equation using L'hôpital theorem. But i put a lot of effort into really understanding what I'm doing, therefore i outlined the fact that it does not make intuitive sense to me. I was hoping that asking the question here would yield some answers, and it sure did. – Aron Nov 16 '16 at 20:25

$\dfrac{\infty}{\infty}$ is a symbol that in this context just means the numerator and denominator both "approach infinity" (grow without bound). It's true that the numerator approaches infinity far slower than the denominator, which is why $\lim_{x\to\infty}}\dfrac{\ln x}{x}=$.

$\dfrac{\infty}{\infty}$ is called an "indeterminate form" because it doesn't tell you enough information to determine what the limit is.

• Alright, i think i understand it more clearly now. $\frac{\infty}{\infty}$ or $\frac{0}{0}$ is just the convention we use to express something that either goes to zero, or infinity for both functions. – Aron Nov 17 '16 at 19:16

Notice that for $x \ge 1$

$$0 \le \frac {\ln x} x \le \frac {\sqrt x} x = \frac 1 {\sqrt x}$$

and the last fraction tends to $0$ when $x \to \infty$, which means that $\frac {\ln x} x \to 0$ too. This shows that even though both $\ln x$ and $x$ tend to $\infty$, your intuition that $\ln x$ is much slower than $x$ is correct. Nevertheless, "evaluating" the fraction $\frac {\ln x} x$ directly in $\infty$ leads to the undefined object $\frac {\ln \infty} \infty = \frac \infty \infty$, therefore in analysis we never "evaluate at infinity", but rather "take the limit at infinity".

• that is squeeze theorem giving $\lim_{x\to \infty}\frac{\ln x}{x}=0$ – Bhaskara-III Nov 18 '16 at 12:01

This is a simple case of an indeterminate form, more specifically, $$\frac{\infty}{\infty}$$ Using L'Hopital's's rule, differentiate the numerator and denominator

$$\lim_{x\to \infty}\frac{\ln x}{x} = \lim_{x\to \infty}\frac{\frac{d}{dx}\ln x}{\frac{d}{dx}x} = \lim_{x\to \infty} \frac{\frac{1}{x}}{1}=\lim_{x\to \infty}\frac{1}{x}=0$$

• Correct, but I believe that if the OP asked this question, then he surely does not know about derivatives. – Alex M. Nov 16 '16 at 19:54
• That can be the case. Nonetheless, this will help in the future once the skill is learned. – dovedevic Nov 16 '16 at 19:55

We use indeterminate forms because it helps us identify particular groups of problems that we can tackle in a singular way. It's a pretty common tool in maths.

Think about derivatives. Maybe you first learned about finding the derivative via "first principles", i.e. evaluating

$\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$

So you do that for some basic polynomials, a couple of trigonometric functions, maybe the exponential or logarithm functions, and you get a feel for how it works. But if you were asked to differentiate $f(x) = \frac{xe^x}{\sin(x^2 - 4)}$, are you going to write

$\lim_{h \rightarrow 0} \frac{\frac{(x+h)e^{x+h}}{\sin((x+h)^2 - 4)} - \frac{xe^x}{\sin(x^2 - 4)}}{h}$

then rearrange it and try to make it something usable? I sincerely hope not. Instead, you'll go "Well there's a fraction of two functions, so I'll use the Quotient Rule, then in the numerator there's two functions multiplied together, so I'll use the Product Rule on that, then in the denominator there's a function of a function so I'll use the Chain Rule". By identifying certain "classes" of functions, you can define rules to apply to find their derivatives.

It's the same with limits. You already know that if you have the quotient of two functions, $\frac{f(x)}{g(x)}$, and you take a limit where they both go to a finite value (and $g(x)$ doesn't go to zero), then you can just substitute those values. Why? Because there's a theorem that if $f(x) \rightarrow a$ and $g(x) \rightarrow b \neq 0$, then $\frac{f(x)}{g(x)} \rightarrow \frac{a}{b}$.

Similarly, if only one of the two functions follows that rule, then there's another theorem you can apply to say where the limit goes. For example, if in a particular limit $f(x) \rightarrow \infty$ and $g(x) \rightarrow b$ (where, remember, a function "approaching infinity" is just shorthand for "grows without bound and hence has no finite limit"), then $\frac{f(x)}{g(x)} \rightarrow \infty$.

Indeterminate forms happen to be a class of limits where you can't apply a single rule, because for any kind of indeterminate form you can find a particular instance that has any limit you choose. For example, in the case of $\frac{\infty}{\infty}$, we have $\lim_{x \rightarrow \infty}\frac{x}{kx} = \frac{1}{k}$, $\lim_{x \rightarrow \infty}\frac{x^2}{kx} = \infty$ and $\lim_{x \rightarrow \infty}\frac{x}{kx^2} = 0$.

So maybe you define a rule if you're working with polynomials, but then what if someone throws a logarithm or exponential function into the mix? Or a trig function? Sure you can say "$\log(x)$ grows slower than $x$", but how do you know? How does $xe^{x^2}$ go against $x^2 e^x$? Does $\sin(e^x) \log(x)$ even do anything sensible?

The nice thing is that we can tackle an extremely broad range of these indeterminate forms with a single rule. So the fact that we can (1) define a class of limits and (2) determine the behaviour of all members of that class with one method is why we use indeterminate forms. If we didn't have l'Hopital's rule we probably wouldn't mention them so much.

 Actually it's a bunch of related rules and they only apply if the functions involved are sufficiently well-defined, but that still happens to be good enough for a wide variety of functions that we actually care about.

Don't worry too much about the $\frac{\infty}{\infty}$ form. It's just a conventional way of saying that the naive approach of taking the limit of the numerator and denominator separately doesn't work in this case (and specifically that it fails because they both go to $\infty$). It doesn't mean that the limit equals $\frac{\infty}{\infty}$ (especially since $\frac{\infty}{\infty}$ isn't actually a number).

You discuss in the question how $x$ grows much faster than $\log(x)$. That intuition is correct, and is precisely why l'Hopital's rule works. Comparing the derivatives of the numerator and denominator is simply the formal way to determine which is increasing more quickly.

(throughout this post, I speak in terms of the extended number line; similar things can be said about the projective number line)

$\infty$ denotes a specific place on the extended number line; any advice you've heard that tells you to think of $\infty$ as something that "grows" is wrong. It stems from the approach of treating $\infty$ as being the limit of something that grows without bound, but has made the mistake of conflating $f$ with $\lim f$.

While the advice to treat $\infty$ as the limit of something that grows without bound is common, I think it's bad advice — e.g. it's practically begging people to make the conceptual mistake I describe above. In my opinion, the main difficulty to working with $\infty$ directly is simply that textbooks avoid doing so, so students don't get a chance to learn how.

That $\ln x$ grows more slowly than $x$ is no obstacle to $\ln(\infty) = \infty$; recall how inequalities behave in regards to limits:

If $f(x) < g(x)$ for all $x$ and both limits exist, then $\lim_{x \to a} f(x) \leq \lim_{x \to a} g(x)$

With the definition the value of $\ln(\infty)$ is $\infty$, the logarithm is continuous at $\infty$, which is why we make that definition. This lets us make lots of calculations quickly by the usual rule of computing limits of continuous functions by plugging in values; e.g.

$$\lim_{x \to \infty} \frac{x \ln(x)}{x + 1} = \lim_{x \to \infty} \frac{\ln(x)}{1 + \frac{1}{x}} = \frac{\ln(\infty)}{1 + \frac{1}{\infty}} = \frac{\infty}{1 + 0} = \infty$$

However, the quotient $\frac{\infty}{\infty}$ is not defined, for similar reasons as to why $\frac{0}{0}$ is not defined. Consequently, when we consider applying the above rule in the problem

$$\lim_{x \to \infty} \frac{\ln(x)}{x} \mathop{=}^? \frac{\ln(\infty)}{\infty}$$

it doesn't work, because the right hand side is not in the domain of division.

That is, to explicitly answer one of your explicit questions, the typical reason we would consider the indeterminate form reason $\frac{\infty}{\infty}$ when computing $\lim_{x \to \infty} \frac{\ln(x)}{x}$ is simply for the mundane reason that we were hoping to apply a quick method for computing limits, and got the indeterminate form was the result which indicates that the quick method doesn't apply.

Yes, $\ln x$ grows slower than $x$. And somehow, intuitively you may guess that it is so much slower than the $x$ will dominate and force the limit to zero. But that's just guesswork and hunches. You can't really know that the limit is zero until you have proved it.

L'Hôpital's Rule is a way of taking your idea about $\ln x$ growing slower and making a rigorous proof out of it. We take the derivatives of the numerator and denominator to see how fast they each grow. And then we prove that the denominator (in this case) outpaces the numerator enough to make the limit be zero.

For all $x\gt0$, we have $\log(x)\le x-1$. Thus, for $x\ge1$, $$\log(x)=2\log\left(x^{1/2}\right)\le2\left(x^{1/2}-1\right)$$ Thus, $$\frac{\log(x)}{x}\le2\left(x^{-1/2}-x^{-1}\right)$$ Therefore, \begin{align} \lim_{x\to\infty}\frac{\log(x)}{x} &\le2\left(\lim_{x\to\infty}x^{-1/2}-\lim_{x\to\infty}x^{-1}\right)\\ &=0 \end{align}

Infinity is not a number, it's an error condition. Saying that a limit "equals infinity" is just a stupid and confusing way of saying that the limit does not exist because the value increases in an unbounded manner. In particular, the "indeterminate form" should be taken behind the shed and shot, because you can't divide infinity by anything, much less infinity. (In a similar fashion, two wrongs don't make a right because addition is not defined over the space of wrongs.)

The fact that we still use this "indeterminate form" notation in undergraduate calculus courses amazes me. There is no similar notation in formal real analysis.

• "There is no similar notation at higher levels of mathematics" is just not correct. The extended reals (and certainly addition and limits involving them) are commonplace in analysis and topology. – 6005 Nov 17 '16 at 3:43
• The limit of $\ln x$ is literally equal to infinity, it's not just a notation. It's the topology on the extended reals. – 6005 Nov 17 '16 at 3:44
• You don't learn calculus over the extended reals in undergraduate courses. It's certainly true that there's no similar notation for real analysis. – Reid Rankin Nov 17 '16 at 3:49
• Yeah, I agree that the way it's done in undergrad courses (at least, calc courses) doesn't treat infinity rigorously, but the part I disagree with is "there's no similar notation in formal real analysis". I think in formal real analysis when we write infinity we're dealing with the extended reals. – 6005 Nov 17 '16 at 3:51
• Yes, that is true. Thanks for clarifying your post, now what you are criticizing is the idea of an "indeterminate form" and your statement is accurate (there is no similar notation in real analysis). – 6005 Nov 17 '16 at 3:59

Mark S. is right -- $\frac{\infty}{\infty}$ is an indeterminate form, so it doesn't tell you enough information to know that the limit is $0$.

As an interesting sidenote, though, you can replace $\infty$ with an infinite hyperreal number to reason informally that $\ln \infty$ is much less than $\infty$, thus the ratio is $0$. Here's how the reasoning goes:

Let $R$ be an infinite hyperreal. Then $$\frac{\ln R}{R} < \epsilon$$ for any finite real $\epsilon > 0$, becuase $\ln x < \epsilon x$ for sufficiently large real $x$. Therefore $\frac{\ln R}{R}$ is an infinitesimal number, i.e. infinitely close to zero. We conclude that the limit $$\lim_{x \to \infty} \frac{\ln x}{x} = 0,$$ because plugging in any infinite $x$ we get an infinitesimal result.

You are asking for an intuitive explanation. The intuitive explanation is as follows.

Your limit seems to be $\frac{\infty}{\infty}$ . The problem is, $\frac{\infty}{\infty}$ is not a single number.

When you divide finite numbers and the denominator is not 0, the result is always known exactly, what number that is.

When you start to admit infinities into the picture, or 0 in the denominator, the result sometimes is not a single number.

In fact, $\frac{\infty}{\infty}$ could be 0, or any positive number, or $\infty$ , or even, it is possible that it does not have a meaningful "value" at all.

So, if you want to know your limit and your method is: take the limit of the numerator and divide by the limit of the denominator, and you get $\frac{\infty}{\infty}$ , it means, you have to find a different method to get the single answer.

Others have commented on various methods that one can use in this case, and in fact it turns out in your case, there is a meaningful answer, and it is 0.

Roughly:$$\frac{\ln{\left(\infty\right)}}{\infty} ~~\approx~~ \lim_{\begin{array}{c} x \to \infty \\[-25px] y \to \infty \end{array}}{\left( \frac{\ln{\left(x\right)}}{y} \right)} \,.$$ While both $$x$$ and $$y$$ are $$ \infty " ,$$ they're not necessarily the same infinity. This lack of clarity makes it "indeterminate".

### Explaining the error in the question.

The problem with $$\frac{\ln{\left(\infty\right)}}{\infty} ~~=~~ \lim_{x \to \infty}{\left( \frac{\ln{\left(x\right)}}{x} \right)} ~~=~~ 0$$ is that it incorrectly assumes that both $$ \infty " \text{'s}$$ are the same value.

Unless you have additional contextual information that constrains both infinities to being the same value. In that case, it's actually true that $$\frac{\ln{\left(\infty\right)}}{\infty} = 0 .$$

That's what we mean by "indeterminate form": the expression, by itself, doesn't have enough information to determine it. But if there's additional information from the external context, that might change things.

### Related example for interdeterminate forms.

Define $$a \equiv 5 .$$ Then multiply by sides by zero and divide by zero to solve: $$a = 5 ~~\Rightarrow~~ 0 \times a = 0 \times 5 ~~\Rightarrow~~ a = \frac{0 \times 5}{0} ~~\Rightarrow~~ a = \frac{0}{0} \,,$$ making $$a = \frac{0}{0} ,$$ which is "indeterminate".

So what's $$ a + 2 " ?$$ Well, $$a + 2 ~~\Rightarrow~~ \frac{0}{0} + 2 ~~\Rightarrow~~ \frac{0 + 0 \times 2}{0} ~~\Rightarrow~~ \frac{0}{0} \,,$$ so $$ a + 2 "$$ is also "indeterminate".

The point of this example is that "indeterminate" just means that the expression, by itself, lacks information. This is, while we can arrive at expressions for $$ a "$$ and $$ a + 2"$$ which're both "indeterminate", that doesn't mitigate the fact that we know that those values are $$ 5 "$$ and $$ 7 ",$$ respectively, from the external context.

That's the same issue you get with infinities and whatnot. For example, we can say $$\infty + \infty ~~\Rightarrow~~ \infty \,,$$ because, while we may not know what the two summed infinite values are, it doesn't really matter – they're going to sum to another infinite value anyway.

By contrast, we can't generally assume $$\infty - \infty ~~\Rightarrow~~ 0 \,,$$ because that's not generally true; one infinite value may differ from another by more than $$0 .$$ Though $$\infty - \infty = 0$$ can be true in certain cases, e.g. $$\lim_{x \to \infty}{\left(x - x\right)} = \infty - \infty = 0 \,,$$ because even though we didn't reduce $$ x - x "$$ to $$ 0 "$$ before folding the limit-operator, we know that the resulting infinities are the same value, allowing us to observe that they do differ by $$0$$ in this particular context.

• Note: I wanted to write this answer in common notation to keep stuff short-'n-simple, so a few of the statements may require inferring the additional information that common notation doesn't make explicit. For example, at one point it's written that$$\frac{0 \times a}{0} \to a \,,$$which wouldn't generally work. However, in that scenario, $ \frac{\boxed{~}}{0} "$ is just using standard notation for the reverse-transform of having multiplied by $ 0 " ,$ which was viable since we'd just defined that zero. Glossed over it to keep things short; I figure those who might see it'll get it. – Nat Jul 27 at 23:53

If you use integration to find the average value of $$y$$ of the function $$y = x^{-1}$$ where $$x$$ is between $$1$$ and infinity, the result evaluates to $$\frac{\ln(\infty)}{\infty - 1}$$ and since the highest value of $$y$$ in this range is $$\frac 1 1 = 1$$, and the lowest value is $$\frac 1 \infty = 0$$, the value of $$\frac {\ln(\infty)}{\infty - 1}$$ lies between $$0$$ and $$1$$ (but not including $$1$$) so $$\frac {\ln(\infty)}{\infty}$$ is less than $$1$$.