作业帮lim(x→0)=（arctanx-x）/ln(1+2x^3)
来源：学生作业帮助网 编辑：作业帮 时间：2024/10/05 16:58:26
x)ըx6@Ģļ
݊{:s4 *5mT_`gC/oy6Y.yhƓ˞|9eubg3?j01Zn&iꁍ|7~[];T
-
Ou?rϧxwS^lԧ 3!E2OfYg Q2hBu$ف m
lim(x→0)=（arctanx-x）/ln(1+2x^3)
lim(x→0)=（arctanx-x）/ln(1+2x^3)
lim(x→0)=（arctanx-x）/ln(1+2x^3)
说明：此题书写错误,应该是“lim(x→0)[(arctanx-x)/ln(1+2x^3)]”.
原式=lim(x->0)[(arctanx-x)'/(ln(1+2x^3))'] (0/0型极限,应用罗比达法则)
=lim(x->0)[-(1+2x^3)/(6(1+x^2))]
=-(1+0)/(6(1+0))
=-1/6.
lim(x→0)=（arctanx-x）/ln(1+2x^3)
lim（下面是x→100 ）arctanx=
lim(x趋近于0+）arctanx=?
lim(x+arctanx)/(x-arctanx) (x→∞)
求lim(x→0)arctanx/x的极限,
求极限解题lim(x→∞)=arctanx/x
一.x---->0时,证明lim(arctanx)/x=1
lim(x趋向0)（x-arctanx）/（x-arcsinx）=?
求极限lim(x→0)sinxsin(1/x);lim(x→∞)(arctanx/x)
当x趋向于0+,lim arctanx/lnx=?
lim(n趋近于0)(arctanx)/x
lim(arctanx/sinx) x－>0
lim(x→0 )(arctanx/x)^(1/x^2)求极限求lim(x→0 )(arctanx/x)^(1/x^2)
lim(x+sinx)/(3x-arctanx)(x趋于无穷）
证明:arctanx和x是等价无穷小量证明:lim(x→0)arctanx/x=1,即证明arctanx和x是等价无穷小量,用洛必达法则作可以吧?这题好像是0/0求极限的类型
lim(x→∞)arctanx/x的极限
lim(x→+∞)(2/π*arctanx)^x求极限
求极限lim(2/π*arctanx)^x x→∞