# 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 Vinayak Srivastava
2 months, 1 week ago

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## Comments

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- 2 months, 1 week ago

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- 2 months, 1 week ago

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- 2 months, 1 week ago

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- 2 months, 1 week ago

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

- 2 months, 1 week ago

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

- 2 months, 1 week ago

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- 2 months, 1 week ago

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

- 2 months, 1 week ago

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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

- 2 months ago

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

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Ok thanks :)

- 2 months ago

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

- 2 months ago

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I just like to use the programmable buttons

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It is L' Hopital.

- 2 months ago

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Ok thanks :)

- 2 months ago

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

- 2 months, 1 week ago

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@Kriti Kamal.in which grade are you

- 2 months, 1 week ago

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I am in class 10th.

- 2 months, 1 week ago

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are you from cbse board

- 2 months, 1 week ago

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Yes. I am from CBSE board

- 2 months, 1 week ago

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

- 2 months, 1 week ago

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

- 2 months, 1 week ago

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$ln 0 = -\infty$

- 2 months, 1 week ago

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

- 2 months, 1 week ago

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

- 2 months, 1 week ago

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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?

- 2 months, 1 week ago

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

- 2 months, 1 week ago

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Yes you can apply. Go to this link

- 2 months, 1 week ago

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

- 2 months, 1 week ago

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@Vinayak Srivastava do my discussion

- 2 months, 1 week ago

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@Vinayak Srivastava in which year did you joined brilliant

- 2 months, 1 week ago

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2020 lol @Vinayak Srivastava

- 2 months, 1 week ago

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hahahahha lol

- 2 months, 1 week ago

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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.

- 2 months, 1 week ago

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Oh, thank you!

- 2 months, 1 week ago

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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

- 2 months, 1 week ago

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Oh, so the limit does not exist?

- 2 months, 1 week ago

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Yes, you're right.

- 2 months, 1 week ago

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Thank you for helping!

- 2 months, 1 week ago

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- 2 months, 1 week ago

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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.

- 2 months, 1 week ago

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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!

- 2 months, 1 week ago

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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.

- 2 months, 1 week ago

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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.

- 2 months, 1 week ago

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this link might help :).

- 2 months, 1 week ago

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Yup, your link says it best:

- 2 months, 1 week ago

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

- 2 months, 1 week ago

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yes sir

- 2 months, 1 week ago

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Ohk, so can I write it as:

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

- 2 months, 1 week ago

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I think this is similar to the k/0.

- 2 months, 1 week ago

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I don't understand, sorry.

- 2 months, 1 week ago

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

- 2 months, 1 week ago

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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.

- 2 months, 1 week ago

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Yeah. Something like this

- 2 months, 1 week ago

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