极限考研真题

这属于0/0的极限题，分子应该是e^0-e^0=1-1=0, 不能变其中一个而另一个不变，使得分子成了e^(x^2)-1或是1-e^(2-2cosx)都是错误的。

此题分子分母极限都是0，所以要用法则解才对，就是分子分母分别求导数法。

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分子用有理化的方法处理，整个函数化简为[xcosx-sinx]/x³×1/[√(1+xcosx)+√(1+sinx)]，前者用洛必达法则求解得-1/3，后者的极限值为函数值1/2，所以整个极限值是-1/6分子的极限本来就是0啊，但是分母的极限也是0，所以相除的极限就没人知道了，于是就构成了不定型求极限的零比零型，零比零型求极限通常的方法这里用不上，直接运用洛必达法则求解。

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