lim(x→π/4) tan(π/4-x)/cot2x 求极限
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/09 15:12:31
x)ըx6|BIb[_bTlcӳy/gNI*'A~
?ٽ$ 13 kakT %D}
C] V\Z`d O?or 0MhlbڐJA䁢p!_\g2 [e
lim(2-x)tanπ/4x
lim (x→0) [tan( π/4 - x )]^(cotx)=?
x→1,求lim(tanπx/4)^tanπx/2求极限,
求lim (x→2)(x-2)tanπx/4的值
lim(x→1) (1-x)tan(π/2)x
lim(x→π/4) tan(π/4-x)/cot2x 求极限
求极限:lim (x→0) [tan( π/4 - x )]^(cotx)=?
lim(x趋向于π/4) sinx-cosx/tan^2 x-1
lim(2-x)^tanπx/2 x趋近1
1.计算lim(1/x2-cot2x) X→0 2 lim(1-x)tanπx/2x→1
(1-x)tan(πx/2),lim 趋向于1,
tan(-x+π/4)
求x→1时lim(2-x)^tan(πx)/2的极限
lim(x→1)(x-1)tanπx/2=?请不要用洛必达法则.
lim(x→1)(1-x^2)tanπx/2
求lim(1-x)tanπx/2 x→1
紧急:求lim tan(e^(x-1)-e^(x^2-1))/(arctanx-π/4),X趋于1;
lim(x趋于π(派)) sin3x/tan5x.lim(x趋于π/4(4分之派)) sinx-cosx/1-cos^x.lim(x趋于0) lntan7x/lntan2x.lim(x趋于1)((1-x)*tan 派x/2.lim(x趋于0)x^2*e^x