作业帮lim(x-0+)x∧(x);用洛必达法则求极限.x的x次放
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/17 16:22:22
lim(x-0+)x∧(x);用洛必达法则求极限.x的x次放
lim(x-0+)x∧(x);用洛必达法则求极限.
x的x次放
lim(x-0+)x∧(x);用洛必达法则求极限.x的x次放
lim(x-0+) x^x = lim(x-0+) e^ (x lnx) = e^{ lim(x-0+) x lnx }
lim(x-0+) x lnx = lim(x-0+) lnx / (1/x)
= lim(x-0+) (1/x) / (-1/x²) = lim(x-0+) -x = 0
=> lim(x-0+) x^x = e^0 = 1
lim xlnx (x->+0)
lim (x->0) 3x/(sinx-x)
lim (x→0)x-sinx/x
lim x->0 x/|x| 等于多少?、
lim(1-x)^(2/x) x->0
lim(x趋向0）(e^x-x^-x)/x
(lim) x/(x-y)
紧急!..微积分..lim (Lnx/x) X->0 LIM是多少啊..LIM (INX/X) x->0
lim （sinxe∧x -x（x+1））/x∧3 x趋向于0
lim(x-0+)x∧(x);用洛必达法则求极限.x的x次放
lim(x+t/x-t)∧x x→无穷大
x趋向0 lim xcotx
lim(x→0) xsinx
6.lim/xtanx x-0
lim(0-∞)/(x+1)
求极限lim(x->0)(x+e^x)^2/x
求极限:lim(x->0) ln(sinx/x)/(x*x)
lim x->0 (sin x-tan x)/sin (x^3)