Why does lim x → 0 − I m ( ln ( x ) e x ) = π?Obviously this is no coincidence. I was thinking maybe this has to do with Euler's formula, but I don't see how the logarithm factors in. Does this even have to do with Euler's Formula?

Question

Why does       lim          x      →              0                  −                          I    m        (          ln              (      x      )              e      x        )    =  π?Obviously this is no coincidence. I was thinking maybe this has to do with Euler&#039;s formula, but I don&#039;t see how the logarithm factors in. Does this even have to do with Euler&#039;s Formula?

amilazamiyn · Accepted Answer

For values of   x along the negative real axis, we can write   x  =      |    x      |        e          i      (      2      ℓ      +      1      )      π       for integer values of   ℓ. Therefore, the multi-valued logarithm is given by  log  ⁡  x  =  log  ⁡  (      |    x      |        e          i      (      2      ℓ      +      1      )      π        )  =  log  ⁡      |    x      |    +  i  (  2  ℓ  +  1  )  π  )Finally,                    lim                  x          →                      0                          −                                          Im              (        log        ⁡        (        x        )                  e                      x                          )            =      (      2      ℓ      +      1      )      π      If we restrict ourselves to the principal branch of the complex logarithm, then                    lim                  x          →                      0                          −                                          Im              (        log        ⁡        (        x        )                  e                      x                          )            =      π

Krish Logan · Accepted Answer

lim          x      →              0        −              ln  ⁡  x      e    x                      =                              e          x                →        1                  lim          x      →              0        −              ln  ⁡  x  =      lim          x      →              0        +              ln  ⁡  (  −  x  )  =  ln  ⁡  (  −  1  )  +      lim          x      →              0        +              ln  ⁡  (  x  )The real part of the final expression is undefined (      lim          x      →              0        +              ln  ⁡  (  x  )  =  −  ∞), and the imaginary value depends on the branch under consideration. The imaginary part of the principal value is   π (because the principal value of   ln  ⁡  (  −  1  )  =  i  π).

Why does lim_(x->0^(-)) ln(x)e^x=pi

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