Evaluate lim x → ∞ ( 1 x + 2 1 x ) x

Question

Evaluate       lim          x      →      ∞                  (              1        x            +              2                              1            x                              )        x

Feriheashrz · Accepted Answer

You have that      e          1      +      x              2                              1            x                              −      x        =      e          1      +      x              (                  2                                    1              x                                      −        1        )            Now recall that      lim          t      →      0                          a        t            −      1        t    =      lim          t      →      0                          e                  t          log          ⁡          a                    −      1        t    =      lim          t      →      0                          e                  t          log          ⁡          a                    −      1              t      log      ⁡      a        ⋅  log  ⁡  aBy putting   u  =  t  log  ⁡  a and noticing that   t  log  ⁡  a  →  0 as   t  →  0, hence      lim          t      →      0                          e                  t          log          ⁡          a                    −      1              t      log      ⁡      a        ⋅  log  ⁡  a  =      lim          u      →      0                          e        u            −      1        u    log  ⁡  a  =  1  ⋅  log  ⁡  a  =  log  ⁡  aSo, in       e          1      +      x              (                  2                                    1              x                                      −        1        )            , putting   z  =      1    x   you have that   z  →  0 as   x  →  ∞ and so      lim          x      →      ∞            e          1      +      x              (                  2                                    1              x                                      −        1        )              =      lim          z      →              0        +                  e          1      +                                    2            z                    −          1                z              =      e          1      +      log      ⁡      2        =  e  ⋅      e          log      ⁡      2        =  2  eAlternatively, notice that since when   x  →  ∞ it is       1    x    +      2          1              /            x        &amp;gt;  0, it is            (              1        x            +              2                  1                      /                    x                    )        x    =  exp  ⁡      [    x    log    ⁡          (              1        x            +              2                  1                      /                    x                    )        ]    =  exp  ⁡      {    x          [              1        x            log      ⁡      2      +      log      ⁡              (        1        +                  1                      x            ⋅                          2                              1                                  /                                x                                                    )            ]        }  Now use that   log  ⁡  (  1  +  t  )  =  t  +  o  (  t  ) as   t  →  0; I&#039;m using   exp  ⁡      [    f    (    x    )    ]    =      e          f      (      x      )       for better formatting.

Shea Stuart · Accepted Answer

Taking the logarithm of the function and setting

t := \frac{1}{x}

, we can use L'Hospital:

lim_{t \to 0} \frac{\log (t + 2^{t})}{t} = lim_{t \to 0} \frac{(1 + 2^{t} \log 2)}{t + 2^{t}} .

= 1 + \log 2

, giving the final answer

2 e

Evaluate <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeXAtom-ORD

Answered question

Answer & Explanation

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