I have to solve this limit without using l'Hospital rule, the answer is 1/2 lim x → 0 tan ⁡ ( x 2 ) ( e 2 x − 1 ) sin ⁡ x

Question

I have to solve this limit without using l&#039;Hospital rule, the answer is 1/2      lim          x      →      0                  tan      ⁡      (              x        2            )              (              e                  2          x                    −      1      )      sin      ⁡      x

Kamora Greer · Accepted Answer

Without knowing what       e    x   is as a Taylor series, we&#039;re going to struggle.       e    x   can also be defined in other ways (e.g. as a limit), but it would be hard to evaluate this limit using those definitions. So let&#039;s take for granted:      e          x        =  1  +  x  +            x      2        2    +            x      3        6    +  ⋯  +            x      n              n      !        +  ⋯Now, let&#039;s substitute x out for 2x:      e          2      x        =  1  +  2  x  +            (      2      x              )        2              2    +            (      2      x              )        3              6    +  ⋯  +            (      2      x              )        n                    n      !        +  ⋯... and subtract 1 so we get the denominator of our limit:      e          2      x        −  1  =  2  x  +            (      2      x              )        2              2    +            (      2      x              )        3              6    +  ⋯  +            (      2      x              )        n                    n      !        +  ⋯It would be great if there was a factor of x I could take out of this, to cancel with the numerator...      e          2      x        −  1  =  x      (    2    +                            2          2                x            2        +                            2          3                          x          2                    6        +    ⋯    +                            2          n                          x                      n            −            1                                      n        !              +    ⋯    )  So my limit is now:      lim          x      →      0            x                  e                  2          x                    −      1        =      lim          x      →      0            x          x              (        2        +                                            2              2                        x                    2                +        ⋯        +                                            2              n                                      x                              n                −                1                                                          n            !                          +        ⋯        )              =      lim          x      →      0            1          2      +                                    2            2                    x                2            +      ⋯      +                                    2            n                                x                          n              −              1                                                n          !                    +      ⋯      As   x  →  0, what happens to each of the terms in the denominator?

I have to solve this limit without using l'Hospital rule, the answer is 1/2 <munder> <mo mov

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