I wanted to prove lim x → ∞ ln ⁡ x x = 0 by the squeeze theorem.

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I wanted to prove       lim          x      →      ∞                  ln      ⁡      x        x    =  0 by the squeeze theorem.

hofyonlines5 · Accepted Answer

ln  ⁡  x  =  2  ln  ⁡      x                  1        2              ≤  2      x   (using the inequality   y  ≥  ln  ⁡  y for   y  &amp;gt;  0 that you obtained.)So it follows that: for all   x  &amp;gt;  0      |              ln      ⁡      x        x        |    ≤  2            x        x  It follows by squeeze theorem that       lim          x      →      ∞            |              ln      ⁡      x        x        |    =  0. Now note that   −      |              ln      ⁡      x        x        |    ≤            ln      ⁡      x        x    ≤      |              ln      ⁡      x        x        |   and hence the result follows by squeeze theorem.

Hailie Blevins · Accepted Answer

Extending the approximation      e    x    ≥  1  +  xinto      e    x    ≥  1  +  x  +            x      2        2    &amp;gt;            (      x      +      1              )        2              2  and      e    x    ≥  1  +  x  +            x      2        2    +            x      3        6    &amp;gt;            (      x      +      1              )        3              6  allows      e    x    &amp;gt;            (      x      +      1              )        n                    n      !      for all   n  ∈      Z  So,  x  ≥  n  ln  ⁡  (  1  +  x  )  −  ln  ⁡  n  !  ≥  n  ln  ⁡  x  −  ln  ⁡  n  !So      x    n    +            ln      ⁡      n      !        n    ≥  ln  ⁡  x      1    n    +            ln      ⁡      n      !              n      x        ≥            ln      ⁡      x        x  As we can make both terms on the LHS arbitrarily small, we are done.

I wanted to prove <munder> <mo form="prefix">lim <mrow class="MJX-TeXAtom-ORD"> x

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