I want to prove that lim x → − 1 − log ⁡ ( 1 + 1 x ) − 1 1 + x = + ∞ .

Question

I want to prove that      lim          x      →      −              1                  −                      log  ⁡  (  1  +      1    x    )  −      1          1      +      x        =  +  ∞  .

Melina Richard · Accepted Answer

Let  u  =  −      1          x      +      1      and then   x  =  −            u      +      1        u  . So  log  ⁡  (  1  +      1    x    )  −      1          1      +      x        =  u  −  log  ⁡  (  u  +  1  )and hence if   x  →  −      1    −  , then   u  →  ∞. For   u  &amp;gt;  6      e    u    =  1  +  u  +            u      2        2    +            u      3              3      !        +  ⋯  &amp;gt;  1  +  u  +            u      2        2    +            u      3              3      !        &amp;gt;      u    2    +  2  u  +  1  =  (  u  +  1      )    2    .So  u  −  log  ⁡  (  u  +  1  )  &amp;gt;  log  ⁡  (  u  +  1  )and hence      lim          x      →      −              1                  −                      log  ⁡  (  1  +      1    x    )  −      1          1      +      x        =      lim          u      →      ∞        (  u  −  ln  ⁡  (  u  +  1  )  )  =  ∞  .

Lena Bell · Accepted Answer

If you set   1  +  x  =  −  u, then   x  =  −  u  −  1 and  1  +      1    x    =  1  −      1          u      +      1        =      u          u      +      1      so your limit becomes      lim          u      →              0        +                        (        log  ⁡  u  −  log  ⁡  (  1  +  u  )  +      1    u              )        =      lim          u      →              0        +                        (                  u      log      ⁡      u      +      1        u    −  log  ⁡  (  1  +  u  )            )      and you can use      lim          u      →              0        +              u  log  ⁡  u  =  0  ,        lim          u      →      0        log  ⁡  (  1  +  u  )  =  0

I want to prove that <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJ

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