问题
解答题
求
|
答案
应用罗比塔法则,lim x→0
=ex-e-x-2x x-sinnx lim x→0
=0.(n≠1)ex-e-x-2 1-ncosnx
当n=1时,lim x→0
=ex-e-x-2x x-sinnx lim x→0
=ex-e-x-2 1-cosx lim x→0
=ex-e-x sinx lim x→0
=2ex-e-x cosx
求
|
应用罗比塔法则,lim x→0
=ex-e-x-2x x-sinnx lim x→0
=0.(n≠1)ex-e-x-2 1-ncosnx
当n=1时,lim x→0
=ex-e-x-2x x-sinnx lim x→0
=ex-e-x-2 1-cosx lim x→0
=ex-e-x sinx lim x→0
=2ex-e-x cosx