Evaluating limit using logarithms. lim x → ∞ ( ln ⁡ x ) 1 x What i have tried: ln ⁡ [ lim x → ∞ ( ln ⁡ x ) 1 x ] lim x → ∞ ln ⁡ ( ln ⁡ x ) 1 x lim x → ∞ ln ⁡ ( ln ⁡ x ) x So as x approaches infinity, the limit goes to 0. But the answer in the book is 1.

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Evaluating limit using logarithms.      lim          x      →      ∞        (  ln  ⁡     x      )                  1        x            What i have tried:  ln  ⁡      [          lim              x        →        ∞              (    ln    ⁡         x          )                        1          x                      ]        lim          x      →      ∞        ln  ⁡  (  ln  ⁡  x      )                  1        x                  lim          x      →      ∞                  ln      ⁡      (      ln      ⁡      x      )        x  So as   x approaches infinity, the limit goes to 0. But the answer in the book is 1.

aqueritztv · Accepted Answer

You took the natural log   ln of the limit to evaluate it easier, but you forgot to undo the natural log. It is just like how if you were to add   1 to the limit to make it easier to calculate, you would have to subtract off   1 in the end.In this case, to &quot;undo&quot; a natural log, you take   e to the power of something. So after you took the natural log you calculated the limit to be   0; then      e    0    =  1

Davin Fields · Accepted Answer

I am still learning about limits, but I think the following approach ought to work.For any   p  ,  q         p    q    =  exp  ⁡  (  ln  ⁡  (  p  )  q  ). Hence  ln  ⁡  (  x      )          1              /            x        =  exp  ⁡      (                  ln        ⁡        (        ln        ⁡        x        )            x        )  And taking limits                              lim                      x            →            ∞                          ln        ⁡        (        x                  )                      1                          /                        x                                              =                  lim                      x            →            ∞                                    (          exp          ⁡                      (                                          ln                ⁡                (                ln                ⁡                x                )                            x                        )                    )                                                  =        exp        ⁡                  (                      lim                          x              →              ∞                                            (                                          ln                ⁡                (                ln                ⁡                x                )                            x                        )                    )                    and hopefully you can see the inner limit evaluates to 0.

Evaluating limit using logarithms. <munder> <mo movablelimits="true" form="prefix">lim

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