I need to find lim x → 0 e x sin ⁡ x − x ( 1 − x ) x 3

Question

I need to find      lim          x      →      0                          e        x            sin      ⁡      x      −      x      (      1      −      x      )              x      3

thoumToofwj · Accepted Answer

Actually,      lim          x      →              0        +                                e        x            cos      ⁡      (      x      )      +      1              3      x        =  ∞  ,and it follows from L&#039;Hopital&#039;s Rule that      lim          x      →              0        +                                e        x            sin      ⁡      (      x      )      −      x      (      1      −      x      )              x      3        =  ∞  .By the same argument,      lim          x      →              0        −                                e        x            sin      ⁡      (      x      )      −      x      (      1      −      x      )              x      3        =  −  ∞  .So, the limit      lim          x      →      0                          e        x            sin      ⁡      (      x      )      −      x      (      1      −      x      )              x      3            lim          x      →      0                          e        x            sin      ⁡      (      x      )      −      x      (      1      −      x      )              x      3      doesn&#039;t exist indeed.

Alaina Marshall · Accepted Answer

As an alternative way, we can avoid l&#039;Hopital&#039;s rule noting that                    e        x            sin      ⁡      x      −      x      (      1      −      x      )              x      3        =                    e        x            sin      ⁡      x      −      sin      ⁡      x      +      sin      ⁡      x      −      x      (      1      −      x      )              x      3        =  =                    e        x            sin      ⁡      x      −      sin      ⁡      x              x      3        +            sin      ⁡      x      −      x      (      1      −      x      )              x      3        =  =      1    x              sin      ⁡      x        x                      e        x            −      1        x    +            sin      ⁡      x      −      x      +              x        2                    x      3        =  =      1    x              sin      ⁡      x        x                      e        x            −      1        x    +            sin      ⁡      x      −      x              x      3        +      1    x  with             sin      ⁡      x        x    →  1,                     e        x            −      1        x    →  1,             sin      ⁡      x      −      x              x      3        →  −      1    6   but limit for       1    x   doesn&#039;t exist.

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