求极限 答案是e的-1次
网友回答
设:y = ((3-e^x)/(x+2))^(1/sin(x))
lny=1/sin(x)*ln((3-e^x)/(x+2));
现求lim(1/sin(x)*ln((3-e^x)/(x+2)))当x->0时极限;对上式进行求导得:
(-e^x*ln(e)/(x+2)-(3-e^x)/(x+2)^2)/cos(x)
把x=0,代入得lny=-1;
故y=e^-1
======以下答案可供参考======
供参考答案1:
利用等价无穷小
ln(1+x) ~ x ,sinx ~ x (x→0),
极限 lim(x→0)(1/sinx)*ln[(3-e^x)/(x+2)]
= lim(x→0)(1/sinx)*ln[1+(1-x-e^x)/(x+2)]
= lim(x→0)(1/sinx)*[(1-x-e^x)/(x+2)]
= lim(x→0)[1/(x+2)]*lim(x→0)[(1-x-e^x)/x]
= (1/2)*lim(x→0)[(1-x-e^x)/x] (0/0)
= (1/2)*lim(x→0)[(-1-e^x)/1]
= (1/2)*(-2)
= -1,
所以 g.e. = e^(-1)。