求极限limx→π+/2ln(x-π/2)/ranx.
这是一个∞/∞型未定式,根据洛必达法则,有 limx→π+/2[ln(x-π/2)]′/(tanx)′=limx→π+/2[1/(x-π/2)]/sec2x=limx→π+/2cos2x/(x-π/2)(0/0型) =limx→π+/2(cos2x)′/(x-π/2)′=limx→π+/2-2cosxsinx/1=0 所以=limx→π+/2[ln(x-π/2)]/tanx=0
求极限limx→π+/2ln(x-π/2)/ranx.
这是一个∞/∞型未定式,根据洛必达法则,有 limx→π+/2[ln(x-π/2)]′/(tanx)′=limx→π+/2[1/(x-π/2)]/sec2x=limx→π+/2cos2x/(x-π/2)(0/0型) =limx→π+/2(cos2x)′/(x-π/2)′=limx→π+/2-2cosxsinx/1=0 所以=limx→π+/2[ln(x-π/2)]/tanx=0