Let L ( X ) = exp ⁡ ( log ⁡ X log ⁡ log ⁡ X )Prove that if c&gt;0, Y = L ( X ) c , and u = log ⁡ X / log ⁡ Y , then u u = L ( X ) ( 1 / 2 c ) ( 1 + o ( 1 ) ) I've tried to write u u = ( log ⁡ X / log ⁡ Y ) log ⁡ X / log ⁡ Y But that doesn't seem to get me anywhere.

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Let   L  (  X  )  =  exp  ⁡  (      log    ⁡    X    log    ⁡    log    ⁡    X    )Prove that if c&amp;gt;0,  Y  =  L  (  X      )    c  , and   u  =  log  ⁡  X      /    log  ⁡  Y , then      u    u    =  L  (  X      )          (      1              /            2      c      )      (      1      +      o      (      1      )      )      I&#039;ve tried to write       u    u    =  (  log  ⁡  X      /    log  ⁡  Y      )          log      ⁡      X              /            log      ⁡      Y      But that doesn&#039;t seem to get me anywhere.

kuthiwenihca · Accepted Answer

First, we have   ln  ⁡      u    u    =  u  ln  ⁡  u  =      1          ln      ⁡      Y            (    ln    ⁡    X    ln    ⁡                  ln        ⁡        X                    ln        ⁡        Y              )    =      1          ln      ⁡      Y            (    ln    ⁡    X    (    ln    ⁡    ln    ⁡    X    −    ln    ⁡    ln    ⁡    Y    )    )  , and   ln  ⁡  ln  ⁡  Y  =  ln  ⁡  (  c  ⋅      ln    ⁡    X    ln    ⁡    ln    ⁡    X    )  =      1    2    ln  ⁡  ln  ⁡  X  +  o  (  ln  ⁡  ln  ⁡  X  )  ,     a  s     X  →  +  ∞  . Therefore,   ln  ⁡      u    u    =      1          2      c              ln        ⁡        X        ln        ⁡        ln        ⁡        X              ln  ⁡  X  ln  ⁡  ln  ⁡  X  ⋅  (  1  +  o  (  1  )  )  . On the other hand, we have   ln  ⁡  L  (  X  )  =      ln    ⁡    X    ln    ⁡    ln    ⁡    X  , and so            ln      ⁡              u        u                    ln      ⁡      L      (      x      )        =      1          2      c        (  1  +  o  (  1  )  )  .

Let L(X)=exp(sqrt(log X log log X)) Prove that if c>0,Y=L(X)^c, and u=log X / log Y , then u^u=L(X)^{(1/2c)(1+o(1))}

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