lim（x→0＋）ln（1－cosx）/lnx

Question

lim（x→0＋）ln（1－cosx）/lnx

Answer 1

∵lim(x->0+)(sinx/x)=1,lim(x->0+)((1-cosx)/x^2)=1
∴lim(x->0+)[ln(1-cosx)/lnx]
=lim(x->0+)[(sinx/(1-cosx))/(1/x)] (∞/∞型极限,应用罗比达法则)
=lim(x->0+)[(sinx/x)/((1-cosx)/x^2)]
=[lim(x->0+)(sinx/x)]/[lim(x->0+)((1-cosx)/x^2)]
=1/1
=1.