Question

limx趋向于0[ln(1+x)]/x^2。用洛必达法则求极限

Answer 1

令y=x^tanx，且两边同时取对数。则lim(x趋于0+)x^tanx=lim(x趋于0+)y=lim(x趋于0+)e^lny=lim(x趋于0+)e^(tanx*lnx)=e^[lim(x趋于0+)tanxlnx].现在分析[lim(x趋于0+)tanxlnx]，=lim(x趋于0+)
[
lnx
/
(1/tanx)
]=lim(x趋于0+)
(1/x)
/
{
[-sec^(2)x]/[tan^(2)x]
}=lim(x趋于0+)
(1/x)/[(-secx/tanx)^2]
=lim(x趋于0+)
(1/x)
/
-{
[1/cosx)/tanx]^2
}=lim(x趋于0+)
(1/x)
/
-{1/[sin^(2)x]
}=lim(x趋于0+)
-sin^(2)x/x=lim(x趋于0+)
(x^2)/x=-x=0，所以待求极限为e^0=1。