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@Pi Han Goh
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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.
It is because computers are fitted with code that x0=1 for all x as codes need to evaluate it seperately. So, it gave the output 1. @Vinayak Srivastava
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.
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=nn. Then lnm=nlnn.
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@Pi Han Goh, @Mahdi Raza, @Aryan Sanghi, @Siddharth Chakravarty
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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.
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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!
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−∞ and if +0, then +∞
If -0, thenLog in to reply
x→0+lim01=+∞ and x→0−lim01=−∞
Oh, so you mean:I also think this is similar, but I don't understand @Pi Han Goh's comment.
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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.
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x→∣0∣limxx?
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this link might help :).
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@Kriti Kamal and @Pi Han Goh :)
Thank youIt is because computers are fitted with code that x0=1 for all x as codes need to evaluate it seperately. So, it gave the output 1. @Vinayak Srivastava
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Oh, so the limit does not exist?
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Yes, you're right.
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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.
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Oh, thank you!
@Vinayak Srivastava do my discussion
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=nn. Then lnm=nlnn.
So, n→0limlnm=n→0limn1lnn.
This is in the form ∞∞
So, applying L'Hospital's rule to this we get
n→0limlnm=0⟹n→0limm=1
That is, n→0limnn=1.
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I have learnt a little bit of this rule. But how is ln0=∞?
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ln0=−∞
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@Vinayak Srivastava,ln(0+h)=-infinite, however ln (0-h) will be undefined .So,the limit doesn't exist.
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ln0=−∞. Also, does LHopital apply in ∞−∞ case?
Oh, the limit is defined. But I don't think we should writeLog in to reply
@Vinayak Srivastava, Limit is not defined because ln(0-h) is not defined. But, ln(0+h)=- infinity.
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Go to this link
Yes you can apply.Sir, I think ln0- will be undefined. So,the limit doesn't exist.
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@Kriti Kamal.in which grade are you
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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
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It is L' Hopital.
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LOL L’Hôpital But in UK many people can't pronounce that, so L'Hospital
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Hey can @Vinayak Srivastava,@Páll Márton,@Siddharth Chakravarty solve https://brilliant.org/problems/an-atm-q/
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@Vinayak Srivastava do
@Vinayak Srivastava . jai shree krishna @Aryan Sanghi.jai shree krishna @Siddharth Chakravarty.jai shree krishna @Kriti Kamal.jai shree krishna
@Kriti Kamal do this
@Vinayak Srivastava.do this
@Aryan Sanghi.do this
@Siddharth Chakravarty.do this
@Vinayak Srivastava in which year did you joined brilliant
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2020 lol @Vinayak Srivastava
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hahahahha lol