是 lim (x ->0+) ln x *ln (x+1)
= = = = = = = = =
令 t =ln (x+1),
則 x =e^t -1,
且 當(dāng) x ->0+ 時,
t ->0+ .
所以 原式=lim (t ->0+) t* ln (e^t -1)
=lim (t ->0+) ln (e^t -1) / (1/t).
由洛必達(dá)法則,
原式=lim (t ->0+) [ e^t /(e^t -1) ] / (-1/t^2)
= -lim (t ->0+) (t^2 *e^t) /(e^t -1)
= -lim (t ->0+) (2t* e^t +t^2 *e^t) /(e^t)
= -lim (t ->0+) (2t +t^2)
= 0.
= = = = = = = = =
解法2 ：因為 當(dāng) x->0+ 時,
ln (1+x) x,
所以 原式=lim (x ->0+) x ln x
=lim (x ->0+) ln x /(1/x).
由洛必達(dá)法則,
原式=lim (x ->0+) [ (1/x) / (-1/x^2) ]
= -lim (x ->0+) x
= 0.