神马作文网〖急〗用洛必达法则求函数极限
来源：学生作业帮 编辑：神马作文网作业帮 分类：数学作业 时间：2024/11/18 19:02:28
〖急〗用洛必达法则求函数极限
9）lim(x→1)[xlnx-(x-1)]/[(x-1)lnx] (0/0)
= lim(x→1)[(lnx+1)-1]/[lnx+(x-1)/x]
= lim(x→1)lnx/(lnx+1-1/x) (0/0)
= lim(x→1)(1/x)/[1/x+1/(x^2)]
= 1/2.
11）先计算
lim(x→0+)sinx*ln|x|
= lim(x→0+)ln|x|/cscx (0/0)
= lim(x→0+)1/x(-cscxcotx)
= -lim(x→0+)(sinx/x)tanx
= 0,
故
g.e.= e^lim(x→0+)sinx*ln|x| = e^0 = 1.
= lim(x→1)[(lnx+1)-1]/[lnx+(x-1)/x]
= lim(x→1)lnx/(lnx+1-1/x) (0/0)
= lim(x→1)(1/x)/[1/x+1/(x^2)]
= 1/2.
11）先计算
lim(x→0+)sinx*ln|x|
= lim(x→0+)ln|x|/cscx (0/0)
= lim(x→0+)1/x(-cscxcotx)
= -lim(x→0+)(sinx/x)tanx
= 0,
故
g.e.= e^lim(x→0+)sinx*ln|x| = e^0 = 1.