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

  1. Sep 29, 2012 #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
     
  2. jcsd
  3. Sep 29, 2012 #2
    Re: limits



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

    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
     
  4. Sep 29, 2012 #3
    Re: limits

    hey the problem is

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

    log n is not in numerator
     
  5. Sep 29, 2012 #4
    Re: limits


    Yeat...too bad you didn't write parentheses in the OP to make that clear, uh?
     
  6. Sep 29, 2012 #5

    micromass

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    Re: limits

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