Question

高数 求极限lim (1+xe^x)^1/sinx x→0

lim (1+xe^x)^1/sinx x→0

Answer 1

原式=lim(1+xe^x)^[(1/xe^x)(xe^x/sinx)] x→0
=e^lim(xe^x/sinx) x→0
=e^lim(xe^x/x) x→0 (sinx与x在x→0时是等价无穷小）
=e^1
=e

Answer 2

lim (1+xe^x)^(1/sinx) x→0 ??
=lim e^{ln[(1+xe^x)^(1/sinx)]} x→0
=lim e^{(1/sinx)·ln(1+xe^x)} x→0
=e^ lim {(1/sinx)·ln(1+xe^x)} x→0
=e^ lim {ln(1+xe^x)/sinx} x→0
=e^ lim {xe^x/x} x→0
=e^ lim {e^x} x→0
=e^1
=e

Answer 3

lim(x->0)(1+xe^x)^(1/sinx)
=lim(x->0)e^[ln(1+xe^x)/sinx]
=e^lim(x->0)ln(1+xe^x)/sinx(LBT法则)
=e^lim(x->0)(x+1)e^x/[cosx(1+xe^x)]
=e