lim(x→0)[cos(1/x)]^x, 先计算  lim(x→0)x*ln[cos(1/x)]  =  lim(t→inf.)(1/t)*ln(cost)  (令  t=1/x)  =  lim(t→inf.)(1/t)*(cost-1)  (等价无穷小替换)  =  ……  =  0, 因此  lim(x→0)[cos(1/x)]^x  =  e^lim(x→0)x*ln[cos(1/x)]  =  e^0  =  1.
limx趋向于无穷(cos1/x)^(x)的极限