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

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  • #1
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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)
cant solve further...please help
 

Answers and Replies

  • #2
606
1


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)
cant 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
 
  • #3
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hey the problem is

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

log n is not in numerator
 
  • #4
606
1


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?
 
  • #5
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This belongs in the homework forum and should contain an effort from the OP. Locked.
 

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