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?

#Calculus

Note by Vinayak Srivastava
10 months, 1 week ago

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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 - 10 months, 1 week ago

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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 - 10 months, 1 week ago

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

A Former Brilliant Member - 10 months, 1 week ago

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@A Former Brilliant Member I don't understand, sorry.

Vinayak Srivastava - 10 months, 1 week ago

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

A Former Brilliant Member - 10 months, 1 week ago

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@A Former Brilliant Member 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 - 10 months, 1 week ago

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@Vinayak Srivastava Yeah. Something like this

A Former Brilliant Member - 10 months, 1 week ago

@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 - 10 months, 1 week ago

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

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

Vinayak Srivastava - 10 months, 1 week ago

@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 - 10 months, 1 week ago

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@A Former Brilliant Member Yup, your link says it best:

Pi Han Goh - 10 months, 1 week ago

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@Pi Han Goh yes sir

SRIJAN Singh - 10 months, 1 week ago

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

Siddharth Chakravarty - 10 months, 1 week ago

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 - 10 months, 1 week ago

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Oh, so the limit does not exist?

Vinayak Srivastava - 10 months, 1 week ago

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Yes, you're right.

Aryan Sanghi - 10 months, 1 week ago

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@Aryan Sanghi Thank you for helping!

Vinayak Srivastava - 10 months, 1 week ago

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 - 10 months, 1 week ago

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Oh, thank you!

Vinayak Srivastava - 10 months, 1 week ago

@Vinayak Srivastava do my discussion

SRIJAN Singh - 10 months, 1 week ago

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.

A Former Brilliant Member - 10 months, 1 week ago

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

Vinayak Srivastava - 10 months, 1 week ago

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ln0=ln 0 = -\infty

A Former Brilliant Member - 10 months, 1 week ago

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@A Former Brilliant Member Its undefined, at least as it is written in my book.

Vinayak Srivastava - 10 months, 1 week ago

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

A Former Brilliant Member - 10 months, 1 week ago

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

A Former Brilliant Member - 10 months, 1 week ago

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@A Former Brilliant Member 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 - 10 months, 1 week ago

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

A Former Brilliant Member - 10 months, 1 week ago

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@A Former Brilliant Member Yes you can apply. Go to this link

A Former Brilliant Member - 10 months, 1 week ago

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

A Former Brilliant Member - 10 months, 1 week ago

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@Kriti Kamal.in which grade are you

SRIJAN Singh - 10 months, 1 week ago

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@Srijan Singh I am in class 10th.

A Former Brilliant Member - 10 months, 1 week ago

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@A Former Brilliant Member are you from cbse board

SRIJAN Singh - 10 months, 1 week ago

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@Srijan Singh Yes. I am from CBSE board

A Former Brilliant Member - 10 months, 1 week ago

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@A Former Brilliant Member I'm in class 7 but how do you know about calculus

SRIJAN Singh - 10 months, 1 week ago

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

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It is L' Hopital.

Siddharth Chakravarty - 10 months ago

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@Siddharth Chakravarty Ok thanks :)

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

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@A Former Brilliant Member 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 - 10 months ago

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@Siddharth Chakravarty I just like to use the programmable buttons

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

SRIJAN Singh - 10 months, 1 week ago

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@Vinayak Srivastava do

SRIJAN Singh - 10 months, 1 week ago

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

SRIJAN Singh - 10 months, 1 week ago

@Kriti Kamal do this

SRIJAN Singh - 10 months ago

@Aryan Sanghi.do this

SRIJAN Singh - 10 months ago

@Vinayak Srivastava in which year did you joined brilliant

SRIJAN Singh - 10 months, 1 week ago

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2020 lol @Vinayak Srivastava

A Former Brilliant Member - 10 months, 1 week ago

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hahahahha lol

SRIJAN Singh - 10 months, 1 week ago
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