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1) 原极限 = lim{x->0} [e^x f'(e^x - 1) - f'(x)]/(3x^2)
= lim{x->0} [e^2x f''(e^x - 1) + e^x f'(e^x - 1) - f''(x)]/(6x)
= lim{x->0} [e^3x f'''(e^x - 1) + e^2x f'(e^x - 1) - f''(x)]/6
= lim{x->0} [e^3x f'''(e^x - 1) - 3]/6 = 3/2, if f'''(0) = 12
2) 原极限 = lim{x->0} {e^[(2/x)ln(1+x)] - e^2[1-ln(1+x)]}/x, simplification first
= lim{x->0} {e^[(2/x)(x-x^2/2+x^3/3] - e^2(1-x+x^2/2)]}/x
= lim{x->0} {e^[(2-x+2x^2/3] - e^2[1-x+x^2/2]}/x
= lim{x->0} e^2{[1-x+2x^2/3+(-x+2x^2/3)^2/2!] - [1-x+x^2/2]}/x
= 0, since the numerator has a higher degree of zero.
= lim{x->0} [e^2x f''(e^x - 1) + e^x f'(e^x - 1) - f''(x)]/(6x)
= lim{x->0} [e^3x f'''(e^x - 1) + e^2x f'(e^x - 1) - f''(x)]/6
= lim{x->0} [e^3x f'''(e^x - 1) - 3]/6 = 3/2, if f'''(0) = 12
2) 原极限 = lim{x->0} {e^[(2/x)ln(1+x)] - e^2[1-ln(1+x)]}/x, simplification first
= lim{x->0} {e^[(2/x)(x-x^2/2+x^3/3] - e^2(1-x+x^2/2)]}/x
= lim{x->0} {e^[(2-x+2x^2/3] - e^2[1-x+x^2/2]}/x
= lim{x->0} e^2{[1-x+2x^2/3+(-x+2x^2/3)^2/2!] - [1-x+x^2/2]}/x
= 0, since the numerator has a higher degree of zero.
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第一题还有其它方法吗
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因为不知道具体的函数形式，想象不到其它方法。
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