题目

lim（x→0）[1-cos（sinx）]/（e^x^2）-1

提问时间：2020-11-04

答案

lim(x-->0) [1 - cos(sinx)]/(e^x² - 1),无穷小替换：sinx~x,e^x² - x,前提是x趋向0
= lim(x-->0) (1 - cosx)/x²
= lim(x-->0) [1 - (1 - 2sin²(x/2))]/x²
= lim(x-->0) [2sin²(x/2)]/(x/2 * 2)²
= lim(x-->0) 2[sin(x/2)]²/(x/2)² * 1/4
= (1/2)lim(x-->0) [sin(x/2)/(x/2)]²,重要定理：lim(x-->0) (sinx)/x = 1
= (1/2)(1)
= 1/2