当x趋向于0时,lim(tanx-sinx)⼀x(sinx)^2极限

2025-01-19 14:16:55
推荐回答(1个)
回答1:

原式=lim(sinx/cosx-sinx)/xsin²x
=lim(1/cosx-1)/xsinx
=lim(1-cosx)/(xsinxcosx)
x趋于0
则1-cosx~x²/2
sinx~x
所以原式=lim(x²/2)/(x²cosx)
=1/2