2(1)等价于limx→1xn−1xm−1=limt→0(t+1)n−1(t+1)m−1=limt→0tn+nt+o(t)tm+mt+o(t)=limt→0tn−1+n+o(1)tm−1+m+o(1)=nm (2)令 1+αx=tm 可得limt→1α(t−1)tm−1=αm (3)limx→01+αxm−1+βxnx=limx→01+αxm−1x−limx→01+βxn−1x=αm−βn (4)原式等价于 ∏i=1nlimx→11−xi1−x=∏i=1n1i=1n! 3(7)在我手上这个版本中是 x→+∞ , 实际上应该是 n→+∞. limn→∞n(cos(xn)−1) 等价于limt→0x2cos(t)−1t2 由于 cos(t)−1∼−12t2 , 结果为 −x22 . 4limn→∞n∑i=1nxin=limn→∞∑i=1nxi=limn→∞x−xn+11−x=x1−x