At what a is lim x → 0 e a x − e x − x x 2 finite? What is that limit?Here is a function: e a x − e x − x x 2 In which a is a coefficient. The problem whats us the value for a which gives a finite value for the limit, lim x → 0 e a x − e x − x x 2 How to find that value? I first though that the degree of the numerator must be equal to that of the denominator. but I don't know what the definition of degree is for e x Even if I knew, I couldn't find the limit.

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At what   a is       lim          x      →      0                          e                  a          x                    −              e        x            −      x              x      2       finite? What is that limit?Here is a function:                    e                  a          x                    −              e        x            −      x              x      2      In which   a is a coefficient. The problem whats us the value for   a which gives a finite value for the limit,      lim          x      →      0                          e                  a          x                    −              e        x            −      x              x      2      How to find that value? I first though that the degree of the numerator must be equal to that of the denominator. but I don&#039;t know what the definition of degree is for       e    x  Even if I knew, I couldn&#039;t find the limit.

Korotnokby · Accepted Answer

Hint. One may recall that, as   u  →  0, by the Taylor series expansion one has,      e    u    =  1  +  u  +            u      2        2    +  O  (      u    3    )  .Can you take it from here?

Leah Pope · Accepted Answer

While Appreciating others&#039; answers, with the help of Olivier Oloa, I post my own solution.      e          a      x        =  1  +  a  x  +            a      2        2        x    2        e          x        =  1  +  x  +      1    2        x    2                      e                  a          x                    −              e        x            −      x              x      2        =            (      a      −      2      )      +      (                        a          2                2            −      1      )      x        x  We want a finite limit, so             a      2        2    −  1  =  1 and   a  −  2  =  0. Which both lead to   a  =  2Evaluating the limit again with known   a,      lim          x      →      0                          e                  2          x                    −              e        x            −      x              x      2        =            1      +      2      x      +      2              x        2            −      1      −      x      −              1        2                    x        2                    x      2        =            (      2      −              1        2            )              x        2                    x      2        =  1.5

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