Hey anyone please can tell me this one:A=lim(e^(1/n*logn)) (n tends

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Discussion Overview

The discussion revolves around evaluating the limit A = lim(e^(1/(n*log(n)))) as n approaches 0. Participants explore different approaches to solving the limit, including the application of logarithms and L'Hôpital's rule, while addressing potential misunderstandings in the formulation of the expression.

Discussion Character

  • Homework-related, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant attempts to solve the limit by taking the logarithm of both sides and applying L'Hôpital's rule, arriving at lim(-1/n) as n approaches 0, but expresses difficulty in proceeding further.
  • Another participant provides a solution using the limit of log(n)/n as n approaches 0, concluding that the limit evaluates to 0, while assuming n is a continuous variable.
  • Some participants clarify that the expression involves e^(1/(n*log(n))) and that log(n) is not in the numerator, suggesting a misunderstanding in the original post.
  • There is a comment regarding the need for clearer notation in the original post to avoid confusion.
  • A participant suggests that the thread belongs in the homework forum and should reflect more effort from the original poster, leading to a lock of the thread.

Areas of Agreement / Disagreement

Participants express differing views on the correct interpretation of the limit and the formulation of the expression. There is no consensus on the resolution of the limit or the appropriateness of the thread's placement.

Contextual Notes

Some limitations include potential misunderstandings in the notation and assumptions regarding the continuity of n. The discussion does not resolve the mathematical steps involved in evaluating the limit.

flash123
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hey anyone please can tell me this one:
A=lim(e^(1/n*logn)) (n tends to 0)
i took log on both sides and then by using l hospital rule
i arrive at lim(-1/n) (n tends to 0)
can't solve further...please help
 
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flash123 said:
hey anyone please can tell me this one:
A=lim(e^(1/n*logn)) (n tends to 0)
i took log on both sides and then by using l hospital rule
i arrive at lim(-1/n) (n tends to 0)
can't solve further...please help



$$\lim_{x\to 0^+}\frac{\log x}{x}=-\infty\Longrightarrow \lim_{n\to 0}e^{\frac{\log n}{n}}=\lim_{x\to -\infty}e^x=0$$

Another way:

$$e^{\frac{\log n}{n}}=\left(e^{\log n}\right)^{1/n}=n^{1/n}\xrightarrow [n\to 0]{} 0$$

Of course, we assume in the above that [itex]\,n\,[/itex] is a continuous variable.

DonAntonio
 


hey the problem is

e^(1/(n*logn)) log n is with n

log n is not in numerator
 


flash123 said:
hey the problem is

e^(1/(n*logn)) log n is with n

log n is not in numerator


Yeat...too bad you didn't write parentheses in the OP to make that clear, uh?
 


This belongs in the homework forum and should contain an effort from the OP. Locked.
 

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