洛必达法则求极限,7 9 11

2025-05-09 08:12:26
推荐回答(1个)
回答1:

(7)原极限=lim(δ->0)e^(1/δln(2-δ))=e^lim(δ->0)(-1/(2-δ)/1)=1/√e
(9)原极限=lim(x->π/2)e^(2cosxlntanx)=e^(2lim(x->π/2)(lntanx/secx))
=e^(2lim(x->π/2)(sec²x/tanx/(secxtanx)))=e^(2lim(x->π/2)(cosx/sin²x)=e^0=1
(11)原极限=lim(x->0+)e^(1/(ln(e^x-1))lnx)=e^(lim(x->0+)(1/x/(1/(e^x-1)*e^x))
=e^(lim(x->0+)(e^x-1)/(xe^x))=e^(lim(x->0+)(e^x/(e^x+xe^x)))
=e^(lim(x->0+)(1/(1+x)=e