Proof lim n → ∞ n ( a 1 n − 1 ) = log ⁡ aI want to show that for all a ∈ R lim n → ∞ n ( a 1 n − 1 ) = log ⁡ aSo far I've got lim n → ∞ n e ( 1 n log ⁡ a ) − n, but when i go on to rearrange this, i come after a few steps back to the beginning...We have no L'Hôpital and no differential and integral calculus so far.

Question

Proof       lim          n      →      ∞        n  (      a                  1        n              −  1  )  =  log  ⁡  aI want to show that for all   a  ∈      R        lim          n      →      ∞        n  (      a                  1        n              −  1  )  =  log  ⁡  aSo far I&#039;ve got       lim          n      →      ∞        n      e          (              1        n            log      ⁡      a      )        −  n, but when i go on to rearrange this, i come after a few steps back to the beginning...We have no L&#039;Hôpital and no differential and integral calculus so far.

Carlos Nicholson · Accepted Answer

Let   x  =  1      /    n. Then you want the limit of                    a        x            −      1        x    =                    e                  x          log          ⁡          (          a          )                    −      1        x    =            1      +      x      log      ⁡      (      a      )      +      R      (      x      log      ⁡      a      )      −      1        x    =  log  ⁡  (  a  )  +  log  ⁡  a      [                  R        (        x        log        ⁡        a        )                    x        log        ⁡        a              ]  Now            R      (      x      log      ⁡      a      )              x      log      ⁡      a        →  0   as   x  →  0Where I have used the fact that      e          θ        =  1  +  θ  +  R  (  θ  )  ,            and       lim          θ      −      &amp;gt;      0                  R      (      θ      )        θ    =  0Here   R  (  θ  ) is   R  (  θ  )  =      ∑          k      =      2        ∞              θ      k              k      !

awezavyoal724 · Accepted Answer

This is essentially the inverse of      e    x    =      lim          n      →      ∞                  (      1      +              x        n            )        n  The sequence of functions       f    n    (  x  )  =            (      1      +              x        n            )        n   converge equicontinuously; simply note that       f    n    ′    (  x  )  =            (      1      +              x        n            )              n      −      1        ∼      e    x  . Thus, we get      lim          n      →      ∞        n      (          e              x                  /                n              −    1    )    =  xwhich is the same as      lim          n      →      ∞        n      (          x              1                  /                n              −    1    )    =  log  ⁡  (  x  )

Proof <munder> <mo form="prefix">lim <mrow class="MJX-TeXAtom-ORD"> n &#

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