ANSWERS: 1

That's an evil one. S=(1+x)^(1/x) can be found using the general binomial expansion: (1+x)^n = 1 + nx + n(n1)x^2/2 + n(n1)(n2)x^3/3! + ... putting in n=(1/x) gets: S=1 + 1 + (1x)/2 + (1x)(12x)/3! + (1x)(12x)(13x)/4! +... Which can be seen to tend to the series for e when x tends to zero. But what is the error? Looking at the coefficient of x gets: C = 0 + 0  1/2  3/3!  6/4!  10/5!  15/6! + ... double that: 2C = 0 + 0  1*2/2  2*3/3!  3*4/4!  4*5/5!  5*6/6! + ... The first two terms of the factorial cancel: 2C = 1  1  1/2!  1/3!  1/4! + ... which is the series for e So S = e  xe/2 + O(x^2) S/e is 1  x/2 + O(x^2) So now the formula is to a first approximation: (1  x/2)^(1/x) Using the binomial expansion AGAIN... (1x/2)^n = 1  nx/2 + n(n1)x^2/2/2^2  n(n1)(n2)x^3/3!/2^3 + ... put n = (1/x) (1+x/2)^(1/x) = 1  1/2 + (1x)/2/2^2  (1x)(2x)/3!/2^3 + ... when x is zero this is 1  1/2 + 1/2/2^2  1/3!/2^3 + 1/4!/2^4 OR 2^0/1  2^1/2 + 2^2/2!  2^3/3! + 2^4/4! +... which I think is e^(1/2) = 1/sqrt(e) Check: evaluate the original equation with x=0.0001 answer 0.60655 Close enough. Phew. There's an easier way if you can use differentiation. Just differentiate the equation on the right and use a single taylor expansion. f'(x) = f(0) + xf'(0) + O(x^2) But differentiation is defined in terms of limits, so that's kind of circular reasoning so I did it all from first principles.
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