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| Forum Index -> Differential Calculus -> View Full Question |
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here it is 1 ^ infinity form thus we write it as = e ^ ([ x]  [0 ] ( f(1+x) / f(1) - 1 ) ^ (1/x) ) = e ^ ([ x] [0 ] ( f(1+x) - f(1)/ f(1) ) ^ (1/x) )=e ^ ([ x] [0 ] ( f(1+x) - f(1)/ x +1 - 1 ) ^ (1/f(1) )= e ^ ([ x] [0 ] ( f ' (1)/ f(1) ) )=e ^ 2
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