Is this limit defined?

I was thinking of this limit which I was curious about, so I typed it in WolframAlpha.

limn0nn\lim \limits_{n\to 0} n^n

I don't think this limit exists, as it should be different from LHL and RHL, but the output came as 11. Why?

Note by Vinayak Srivastava
2 months, 1 week ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

@Pi Han Goh, @Mahdi Raza, @Aryan Sanghi, @Siddharth Chakravarty

Vinayak Srivastava - 2 months, 1 week ago

Log in to reply

If you only consider the real limit (xx is strictly real), then the limit does not exist because LHL does not exist.

But if you consider the complex limit, then the limit does exists!

WolframAlpha considers the latter limit.

Pi Han Goh - 2 months, 1 week ago

Log in to reply

I tried approaching values from both sides.

From the 0+0^+ side, it approaches 11, but from the 00^- side, it approaches 1-1.

Can you please explain what is the difference in real and complex limit? I saw these terms for the first time. Thanks!

Vinayak Srivastava - 2 months, 1 week ago

Log in to reply

@Vinayak Srivastava I think this is similar to the k/0.

Páll Márton (no activity) - 2 months, 1 week ago

Log in to reply

@Páll Márton (no activity) I don't understand, sorry.

Vinayak Srivastava - 2 months, 1 week ago

Log in to reply

@Vinayak Srivastava If -0, then -\infty and if +0, then ++\infty

Páll Márton (no activity) - 2 months, 1 week ago

Log in to reply

@Páll Márton (no activity) Oh, so you mean: limx0+10=+\lim \limits_{x \to {0^{+}}} \dfrac{1}{0}=+\infty and limx010=\lim \limits_{x \to {0^{-}}} \dfrac{1}{0}=-\infty

I also think this is similar, but I don't understand @Pi Han Goh's comment.

Vinayak Srivastava - 2 months, 1 week ago

Log in to reply

@Vinayak Srivastava Yeah. Something like this

Páll Márton (no activity) - 2 months, 1 week ago

Log in to reply

@Vinayak Srivastava real limit is as you've described it: the value of x approaches 0.

for 0^-, the LHL (real) limit does not exist, because x^x is not continuous for non-negative x.

complex limit: the magnitude of x approaches 0.

Pi Han Goh - 2 months, 1 week ago

Log in to reply

@Pi Han Goh Ohk, so can I write it as:

limx0xx?\lim \limits_{x \to |0|} x^x ?

Vinayak Srivastava - 2 months, 1 week ago

Log in to reply

@Pi Han Goh I didn't get the definition of a complex limit, what do you mean by magnitude, the absolute value? If you mean absolute value, then it's the same thing because negative real numbers' absolute value becomes positive and then also approaches 0, so we again obtain the same thing, that the limit does not exist.

Siddharth Chakravarty - 2 months, 1 week ago

Log in to reply

@Siddharth Chakravarty this link might help :).

Kriti Kamal - 2 months, 1 week ago

Log in to reply

@Kriti Kamal Yup, your link says it best:

Pi Han Goh - 2 months, 1 week ago

Log in to reply

@Pi Han Goh yes sir

SRIJAN Singh - 2 months, 1 week ago

Log in to reply

@Pi Han Goh Thank you @Kriti Kamal and @Pi Han Goh :)

Siddharth Chakravarty - 2 months, 1 week ago

Log in to reply

It is because computers are fitted with code that x0=1x^0 = 1 for all xx as codes need to evaluate it seperately. So, it gave the output 11. @Vinayak Srivastava

Aryan Sanghi - 2 months, 1 week ago

Log in to reply

Oh, so the limit does not exist?

Vinayak Srivastava - 2 months, 1 week ago

Log in to reply

Yes, you're right.

Aryan Sanghi - 2 months, 1 week ago

Log in to reply

@Aryan Sanghi Thank you for helping!

Vinayak Srivastava - 2 months, 1 week ago

Log in to reply

The limit does not exist. Because you already gave the reason in your note. Some debate 0^0 is 1, while some have other answers like not defined or in some contexts, it might be in the interminate form depending on what problem it is. To maintain continuity, 0^0 because other numbers like 2^0= 1, and thus is considered 1 in computers.

Siddharth Chakravarty - 2 months, 1 week ago

Log in to reply

Oh, thank you!

Vinayak Srivastava - 2 months, 1 week ago

Log in to reply

@Vinayak Srivastava do my discussion

SRIJAN Singh - 2 months, 1 week ago

Log in to reply

This question can be answered using calculus. You have to use L'Hospital's rule. If you haven't learnt it yet, it will be difficult to understand that. Let m=nnm=n^n. Then lnm=nlnn\ln m=n\ln n.

So, limn0lnm=limn0lnn1n\displaystyle \lim_{n\to 0} \ln m=\displaystyle \lim_{n\to 0} \dfrac {\ln n}{\frac 1n}.

This is in the form \dfrac {\infty }{\infty }

So, applying L'Hospital's rule to this we get

limn0lnm=0    limn0m=1\displaystyle \lim_{n\to 0} \ln m=0\implies \displaystyle \lim_{n\to 0} m=1

That is, limn0nn=1\displaystyle \lim_{n\to 0} n^n=1.

Foolish Learner - 2 months, 1 week ago

Log in to reply

I have learnt a little bit of this rule. But how is ln0=? \ln 0 = \infty?

Vinayak Srivastava - 2 months, 1 week ago

Log in to reply

ln0=ln 0 = -\infty

Páll Márton (no activity) - 2 months, 1 week ago

Log in to reply

@Páll Márton (no activity) Its undefined, at least as it is written in my book.

Vinayak Srivastava - 2 months, 1 week ago

Log in to reply

@Vinayak Srivastava @Vinayak Srivastava,ln(0+h)=-infinite, however ln (0-h) will be undefined .So,the limit doesn't exist.

Kriti Kamal - 2 months, 1 week ago

Log in to reply

@Vinayak Srivastava ln 0 is undefined. But ln (0+h)=- infinity

Kriti Kamal - 2 months, 1 week ago

Log in to reply

@Kriti Kamal Oh, the limit is defined. But I don't think we should write ln0=\ln 0 = -\infty. Also, does LHopital apply in \dfrac{-\infty}{\infty} case?

Vinayak Srivastava - 2 months, 1 week ago

Log in to reply

@Vinayak Srivastava @Vinayak Srivastava, Limit is not defined because ln(0-h) is not defined. But, ln(0+h)=- infinity.

Kriti Kamal - 2 months, 1 week ago

Log in to reply

@Kriti Kamal Yes you can apply. Go to this link

Kriti Kamal - 2 months, 1 week ago

Log in to reply

Sir, I think ln0- will be undefined. So,the limit doesn't exist.

Kriti Kamal - 2 months, 1 week ago

Log in to reply

@Kriti Kamal.in which grade are you

SRIJAN Singh - 2 months, 1 week ago

Log in to reply

@Srijan Singh I am in class 10th.

Kriti Kamal - 2 months, 1 week ago

Log in to reply

@Kriti Kamal are you from cbse board

SRIJAN Singh - 2 months, 1 week ago

Log in to reply

@Srijan Singh Yes. I am from CBSE board

Kriti Kamal - 2 months, 1 week ago

Log in to reply

@Kriti Kamal I'm in class 7 but how do you know about calculus

SRIJAN Singh - 2 months, 1 week ago

Log in to reply

Is it L'Hopital's rule or L'Hospitals rule? I still don't get the spelling....is it Hopital or Hospital? @Foolish Learner @Vinayak Srivastava @Páll Márton

Percy Jackson - 2 months ago

Log in to reply

It is L' Hopital.

Siddharth Chakravarty - 2 months ago

Log in to reply

@Siddharth Chakravarty Ok thanks :)

Percy Jackson - 2 months ago

Log in to reply

LOL L’Hôpital But in UK many people can't pronounce that, so L'Hospital

Log in to reply

@Páll Márton (no activity) I still doubt why do you use so many LOLs, maybe you like to laugh a lot or you are addicted to writing it.

Siddharth Chakravarty - 2 months ago

Log in to reply

@Siddharth Chakravarty I just like to use the programmable buttons

Log in to reply

@Páll Márton (no activity) Ok thanks :)

Percy Jackson - 2 months ago

Log in to reply

Hey can @Vinayak Srivastava,@Páll Márton,@Siddharth Chakravarty solve https://brilliant.org/problems/an-atm-q/

SRIJAN Singh - 2 months, 1 week ago

Log in to reply

@Vinayak Srivastava do

SRIJAN Singh - 2 months, 1 week ago

Log in to reply

@Vinayak Srivastava . jai shree krishna @Aryan Sanghi.jai shree krishna @Siddharth Chakravarty.jai shree krishna @Kriti Kamal.jai shree krishna

SRIJAN Singh - 2 months, 1 week ago

Log in to reply

@Kriti Kamal do this

SRIJAN Singh - 2 months, 1 week ago

Log in to reply

@Vinayak Srivastava.do this

SRIJAN Singh - 2 months, 1 week ago

Log in to reply

@Aryan Sanghi.do this

SRIJAN Singh - 2 months, 1 week ago

Log in to reply

@Siddharth Chakravarty.do this

SRIJAN Singh - 2 months, 1 week ago

Log in to reply

@Vinayak Srivastava in which year did you joined brilliant

SRIJAN Singh - 2 months, 1 week ago

Log in to reply

2020 lol @Vinayak Srivastava

Páll Márton (no activity) - 2 months, 1 week ago

Log in to reply

hahahahha lol

SRIJAN Singh - 2 months, 1 week ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...