lim(sin3x)^1/(1+3lnx) x→0+ 的极限
1个回答

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

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