If therefore we choose a quantity e such that log e I o X X= I, log i oe = X, which gives (by more accurate calculation) e=2.71828..., we shall have lim(loge(I+0))}/0=I, and conversely 'lim' {ex+0 - e x } 143= The deduction of the expansions log e (' +x) = x - Zx 2 + 3x 3 - ..., e x = I +.x+x2/2!+x3
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