# Is this limit defined?

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

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

I don't think this limit exists, as it should be different from LHL and RHL, but the output came as $1$. Why?

Note by Vin_Math 91
10 months, 2 weeks ago

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- 10 months, 2 weeks ago

If you only consider the real limit ($x$ 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.

- 10 months, 2 weeks ago

I tried approaching values from both sides.

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

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

- 10 months, 2 weeks ago

I think this is similar to the k/0.

- 10 months, 2 weeks ago

I don't understand, sorry.

- 10 months, 2 weeks ago

If -0, then $-\infty$ and if +0, then $+\infty$

- 10 months, 2 weeks ago

Oh, so you mean: $\lim \limits_{x \to {0^{+}}} \dfrac{1}{0}=+\infty$ and $\lim \limits_{x \to {0^{-}}} \dfrac{1}{0}=-\infty$

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

- 10 months, 2 weeks ago

Yeah. Something like this

- 10 months, 2 weeks ago

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.

- 10 months, 1 week ago

Ohk, so can I write it as:

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

- 10 months, 1 week ago

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.

- 10 months, 1 week ago

- 10 months, 1 week ago

yes sir

- 10 months, 1 week ago

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

- 10 months, 1 week ago

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

- 10 months, 2 weeks ago

Oh, so the limit does not exist?

- 10 months, 2 weeks ago

Yes, you're right.

- 10 months, 2 weeks ago

Thank you for helping!

- 10 months, 2 weeks 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.

- 10 months, 2 weeks ago

Oh, thank you!

- 10 months, 2 weeks ago

@Vinayak Srivastava do my discussion

- 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 $m=n^n$. Then $\ln m=n\ln n$.

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

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

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

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

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

- 10 months, 1 week ago

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

- 10 months, 1 week ago

$ln 0 = -\infty$

- 10 months, 1 week ago

Its undefined, at least as it is written in my book.

- 10 months, 1 week ago

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

- 10 months, 1 week ago

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

- 10 months, 1 week ago

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

- 10 months, 1 week ago

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

- 10 months, 1 week ago

Yes you can apply. Go to this link

- 10 months, 1 week ago

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

- 10 months, 1 week ago

@Kriti Kamal.in which grade are you

- 10 months, 1 week ago

I am in class 10th.

- 10 months, 1 week ago

are you from cbse board

- 10 months, 1 week ago

Yes. I am from CBSE board

- 10 months, 1 week ago

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

- 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

- 10 months, 1 week ago

It is L' Hopital.

- 10 months, 1 week ago

Ok thanks :)

- 10 months, 1 week ago

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

- 10 months, 1 week ago

I still doubt why do you use so many LOLs, maybe you like to laugh a lot or you are addicted to writing it.

- 10 months, 1 week ago

I just like to use the programmable buttons

- 10 months, 1 week ago

Ok thanks :)

- 10 months, 1 week ago

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

- 10 months, 1 week ago

- 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

- 10 months, 1 week ago

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- 10 months, 1 week ago

@Vinayak Srivastava in which year did you joined brilliant

- 10 months, 1 week ago

2020 lol @Vinayak Srivastava

- 10 months, 1 week ago

hahahahha lol

- 10 months, 1 week ago