Calculate lim x → 0 + 4 x − 1 − x x 2 ln ⁡ ( 1 + e 1 / x ) using L'Hospital's rule.

Question

Calculate       lim          x      →              0                  +                                        4        x            −              1        −        x                            x        2            ln      ⁡      (      1      +              e                  1                      /                    x                    )       using L&#039;Hospital&#039;s rule.

Mya Hurst · Accepted Answer

Your solution is correct. Alternatively, this is how you could proceed without having to use L&#039;Hospital&#039;s rule:                    4        x            −              1        −        x                            x        2            ln      ⁡      (      1      +              e                  1                      /                    x                    )        =      (                            4          x                −        1            x        +                            1          −                      1            −            x                          x              )        1          x      ln      ⁡      (      1      +              e                  1          x                    )        =      (                            4          x                −        1            x        +                  1                  1          +                      1            −            x                                )        1          x      ln      ⁡      (      1      +              e                  1          x                    )      Note that the standard limit                     4        x            −      1        x    →  log  ⁡  4 and that       1          x      ln      ⁡      (      1      +              e                  1          x                    )        =      1          1      +      x      ln      ⁡              (        1        +                  1                      e                          1              x                                      )              →  1 as   x  →  0  + (because       lim          x      →      0      +            1          e              1        x              =  0 hence the denominator has limit   1  +  0  =  1So the required limit is:   log  ⁡  4  +      1    2

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