limx→0 (cosx)^1⼀x 洛必达法则求极限

2025-03-06 12:13:05
推荐回答(2个)
回答1:

原式=e^[lim(x->0)(lncosx)/x]
=e^[lim(x->0)(1/cosx ×(-sinx))/1]
=e^[lim(x->0)-tanx]
=e^0
=1

回答2:

(cosx)^1/x ?
就是(cosx)/x ?
lim (cosx)/x=lim(-sinx)/1|x=0 =0