题目

求极限：lim(x→无穷）（2-x/3-x)的x+2次方

提问时间：2020-07-31

答案

lim(x→∞) [(2-x)/(3-x)]^(x+2)
=lim(x→∞) [(3-x-1)/(3-x)]^(x+2)
=lim(x→∞) [1-1/(3-x)]^(x+2),之后根据e的定义lim(x→∞) (1+1/x)^x,将原式凑成这形式
=lim(x→∞) [1-1/(3-x)]^[-(3-x)]*[-(x+2)/(3-x)],留意指数那是-(3-x),有负号的
=e^-lim(x→∞) (x+2)/(3-x),lim(x→∞) [1-1/(3-x)]^[-(3-x)]这个等于e,之后对指数部分求极限
=e^-lim(x→∞) (1+2/x)/(3/x-1),分子分母各除以x
=e^-(1+0)/(0-1)
=e