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