lim(x->π/2+)[ln(x-π/2)/tanx]=lim(x->π/2+){[1/(x-π/2)]/sec²x} (∞/∞型极限,应用罗比达法则)
=lim(x->π/2+)[cos²x/(x-π/2)] (0/0型极限,应用罗比达法则)
=lim(x->π/2+)(-2sinx*cosx)
=lim(x->π/2+)[-sin(2x)]
=-sinπ
=0.