Limits Strike

Calculus Level 3

$\large \lim_{x\to 0^+} x = 0 \qquad \qquad \lim_{x\to0^+} x^x = 1$

The two limits above are commonly used in calculus textbooks. However, the following limit isn't! What is the value of this limit?

$\large \lim_{x\to0^+} x^{x^x}$

-1

\infty

1

2

0

2 solutions

Pranshu Gaba
Feb 22, 2017

We want to evaluate the limit $\displaystyle \lim _{x \to 0^+} x^{x^x}$ . The correct way to evaluate exponent towers is from top to bottom, so it is better written as $\displaystyle \lim _{x \to 0^+} x^{(x^x)}$ . As $x$ tends to $0$ from the positive side, the base of this exponent tends to 0, and the power tends to 1. This is of the form $0^1$ .

If the limit were of the form $0^0, \, \infty^0,$ or $1^\infty$ , then it would have been an indeterminate form. In these cases, there is a race between the base and the power and we can’t directly determine who will win.

In this question, the limit is of the form $0^1$ which is not a indeterminate form. We can substitute limiting values directly. The statement $\displaystyle \lim _{x\to 0} f(x)^{g(x)} = \displaystyle \lim _{x\to 0} f(x)^{\lim_{x\to 0} g(x)}$ holds true.

$\large \begin{aligned}\lim _{x \to 0^+} x^{(x^x)} &= \lim _{x \to 0^+} x^{ \left(\lim _{x \to 0^+} x^x\right)} \\ & = \lim _{x \to 0^+} x^1 \\ & = 0\end{aligned}$

The base is approaching $0$ , the power is approaches 1, so the exponent as a whole approaches $0^1 = 0$ .

Moderator note:

It's great that you checked the conditions of the theorem before applying in. In this case, to calculate the limit of $f(x) ^ { g(x) }$ under the assumption that $f(x) > 0$ , we proceed by taking logarithms and look at $g(x) \times \ln f(x)$ , so in order for us to apply the product theorem, we have to ensure that these terms do not tend to infinity while the other tends to 0.

$\lim \ln f(x) = - \infty$ (thus $\lim f(x) = 0$ ) and $\lim g(x)= 0$ gives us the $0^0$ indeterminate form.
$\lim \ln f(x) = 0$ (thus $\lim f(x) = 1$ ) and $\lim g(x)= \infty$ gives us the $1^\infty$ indeterminate form.
$\lim \ln f(x) = \infty$ (thus $\lim f(x) = \infty$ ) and $\lim g(x)= 0$ gives us the $\infty ^ 0$ indeterminate form.

This yields the indeterminate forms as listed out, showing how the "validity of product rule" and "validity of exponential rule" are related by taking logarithms.

I must point out that, while the assumption is explicit in the question, the assertion that $x^{x} = 0$ is a convention that, while experiencing recent acceptance (for good reasons), is still a convention that is not--and most definitely has not been--without dispute.

Thus, it might help to expound upon the reasoning behind the indeterminate forms classification?

Joshua Nesseth - 4 years, 3 months ago

Let me clarify. You expressed a common misconception in the phrasing, so bear with the tedious / pedantic tracking of language used.

Note that the distinction between an indeterminate form and an explicit term, is that the indeterminate form refers to the limit of functions/sequences, while the explicit term is just one term.
I disagree with " $\lim_{x \to 0^{+}} x^{x} = 0$ is a convention". It is not defined by convention. The specific limit $\lim_{x \rightarrow 0^+} x^x =0$ can be easily proven through many ways, like by first principles. (It's not been proven in this problem/solution as yet.)
What is defined by convention though, is then when the explicit term $0^0$ is used in a calculation, then people either say it is undefined (Brilliant currently takes this stance), or it is 1 (because it simplifies various calculations where it can be interpreted as the limit of linear/polynomial functions).
Exponding on the reasoning why $0^0$ is an indeterminate form, If $\lim f(x) = 0, \lim g(x) = 0$ , then it is not always true that $\lim f(x) ^ { g(x) } = 1$ . It holds true if both functions were linear/polynomial, but need not be true otherwise .
Similarly, one can create counterexamples for the other forms. Note that there are many paths that allows one to tend to 0, 1, $\infty$ . In most cases, taking a linear / polynomial path leads to the naive expected answer, so to find a counter example we have to get creative (like above).

Calvin Lin Staff - 4 years, 3 months ago

The language is actually appreciated.

1) I was actually speaking more to the classification of $\lim g(x)$ being determined by the $\lim ln f(x)$ .

2) I fully welcome a proof of $\lim_{x \to 0^+ } \, x^{x} = 0$ from First Principles. [Not that I dispute such a statement; I misspoke in my original post--the intent was to speak to the calculable value, not it's limit]

Moreover, I was not intending to assert my own opinions regarding the subject--I am by no means qualified to do so. I was merely pointing out that, given the fact far greater minds than my own have grappled with the concept, that a discussion on such a topic might be beneficial both to students and to the discussion as a whole.

Joshua Nesseth - 4 years, 3 months ago

@Joshua Nesseth – Ah ic. I've cleaned up the phrasing. It's not that the classification of $g(x)$ is dependent on $\ln f(x)$ . It's that if one is $\infty$ and the other is $0$ , then the product rule fails, which explains why these are the indeterminate forms. (IE it's not that we plucked them out from nowhere)

@Pranshu Gaba Note: $0 ^ \infty$ isn't an indeterminate form. It is going to be 0 (once we account for complex exponentiation).

Calvin Lin Staff - 4 years, 3 months ago

@Calvin Lin – Right, I have edited the solution. If the base is very small, then raising it to a very large power will take it zero very quickly. $0^\infty$ is not an indeterminate case.

Pranshu Gaba - 4 years, 3 months ago

Quick question: I was trying to manipulate the problem using logarithms and solve it that way. Something like this (sorry, I'm not very savvy with inserting equations): x^(x^x )= e^ln⁡〖x^(x^x )〗 = e^((x^x)*ln⁡x).

This seems to indicate that the limit would go to 'e'. I'm curious where this path goes wrong. Again, sorry for the poor formatting.

William Venden - 2 years, 7 months ago

Hi William, since $x$ is positive, we can find the limit by taking logarithms. You have written $\large x^{(x^{x})} = e^{\ln(x^{(x^x)})} = e^{(x^x \ln x)}$ What happens to $x^x \ln x$ as $x \to 0^+$ ? Try finding the limits of $x^x$ and $\ln x$ as $x$ goes to zero.

Pranshu Gaba - 2 years, 7 months ago

Whoops, sorry, for some reason I was thinking that ln(0) was 1. My apologies. Makes sense now. Thank you!

William Venden - 2 years, 7 months ago

Samyok Nepal
Feb 21, 2017

Seeing as the second one equals 1, we can plug this into the desired formula to get $\large \lim_{x\to0^+} x^1$

This obviously simplifies into the first limit, therefore the desired limit is equal to $\boxed{0}$ .

That a common misconception in calculus. Do you know why?

Calvin Lin Staff - 4 years, 3 months ago

@Calvin Lin Why ?Also can u tell how limit x-->0+ x^x=1

Kaustubh Miglani - 4 years, 3 months ago

What? And Why?

Aniket Sanghi - 4 years, 3 months ago

More accurately, you should explain that the conditions of the theorem are satisfied, which is why you can separately evaluate the limits. See Pranshu's solution for the details.

Check out Advanced Problem 1 for a case where we have an indeterminate form.

Calvin Lin Staff - 4 years, 3 months ago

(x^x)^x=x^(x^2)

spy koni - 4 years, 3 months ago

Yes, that statement is true.

How does it relate to the problem? Do you remember how exponent towers are evaluated ? We want to look at (x^(x^x)) and not x^(x^x).

Looking at this gave me an idea for a problem :)

Calvin Lin Staff - 4 years, 3 months ago

Limits Strike

2 solutions

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