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lim（x→0＋）ln（1－cosx）/lnx
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lim（x→0＋）ln（1－cosx）/lnx
▼优质解答
答案和解析
∵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.
∴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.
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