lim(sin3x)^1/(1+3lnx) x→0+ 的极限
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/09/21 08:25:09
lim(sin3x)^1/(1+3lnx) x→0+ 的极限
先用洛必达法则:
lim[x→0+] (sin3x)^[1/(1+3lnx)]
=e^lim[x→0+] [1/(1+3lnx)]ln(sin3x)
=e^lim[x→0+] ln(sin3x)/(1+3lnx)
=e^lim[x→0+] (3cos3x/sin3x)/(3/x),上下求导
=e^lim[x→0+] 3cos3x/sin3x·x/3
=e^lim[x→0+] xcos3x/sin3x
=e^lim[x→0+] (cos3x-3xsin3x)/(3cos3x),上下求导
=e^lim[x→0+] (1/3-xtan3x),不为0/0形式,代入数值
=e^(1/3-0)
=e^(1/3)
=e的3次开方根号
lim[x→0+] (sin3x)^[1/(1+3lnx)]
=e^lim[x→0+] [1/(1+3lnx)]ln(sin3x)
=e^lim[x→0+] ln(sin3x)/(1+3lnx)
=e^lim[x→0+] (3cos3x/sin3x)/(3/x),上下求导
=e^lim[x→0+] 3cos3x/sin3x·x/3
=e^lim[x→0+] xcos3x/sin3x
=e^lim[x→0+] (cos3x-3xsin3x)/(3cos3x),上下求导
=e^lim[x→0+] (1/3-xtan3x),不为0/0形式,代入数值
=e^(1/3-0)
=e^(1/3)
=e的3次开方根号
lim(sin3x)^1/(1+3lnx) x→0+ 的极限
求lim(x→0) (1+x)^lnx的极限!
求lim(x→0)x/sin3x的极限
求极限lim(x→0) 3x的平方-5x/sin3x
1.计算下列极限:lim(下面是x→0) (-sin3x+cosx)^(1/x)
用洛必达法则求极限:1、lim(x→0)[e^x-e^(-x)]/sinx 3、lim(x→n)sin3x/tan5x
求极限lim(x→0)sin3x/x
极限Lim(x->0)(sin3x)/x=
求极限lim x→π(sin3x)/(x-π)和求极限lim x→π/2(1+cosx)secx
当x趋于0 lim sin3x/tan7x的极限是多少
求极限lim x→0 sin2x/sin3x ,
x→0 lim sin3x/tan3x 求极限