先用洛必達(dá)法則:
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),上下求導(dǎo)
=e^lim[x→0+] 3cos3x/sin3x·x/3
=e^lim[x→0+] xcos3x/sin3x
=e^lim[x→0+] (cos3x-3xsin3x)/(3cos3x),上下求導(dǎo)
=e^lim[x→0+] (1/3-xtan3x),不為0/0形式,代入數(shù)值
=e^(1/3-0)
=e^(1/3)
=e的3次開(kāi)方根號(hào)