lim(x→1) lncos(x-1)/(1-sin(πx/2))

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lim(x→1)lncos(x-1)/(1-sin(πx/2))lim(x→1)lncos(x-1)/(1-sin(πx/2))lim(x→1)lncos(x-1)/(1-sin(πx/2))原式=l

lim(x→1) lncos(x-1)/(1-sin(πx/2))
lim(x→1) lncos(x-1)/(1-sin(πx/2))

lim(x→1) lncos(x-1)/(1-sin(πx/2))
原式=lim(x→1) ln[1+cos(x-1)-1]/(1-sin(πx/2))
=lim(x→1) [cos(x-1)-1]/(1-sin(πx/2))
=lim(x→1)[-1/2(x-1)^2]/(1-sin(nx/2)
接下就完全是利用洛比达法则了,鉴于符号太难打了,就不赘述了

用洛比达法则