To determine

To evaluate: The limit limx→π2−secxtanx

Answer

The value of the limit is limx→π2−secxtanx=1

Explanation

The given limit is in indeterminate form, we apply l’Hospital’s Rule

 We Know that ddx(secx)=secxtanxand ddx(tanx)=sec2x

limx→π2−secxtanx=limx→π2−secxtanxsec2x=limx→π2−tanxsecx=limx→π2−sec2xsecxtanx=limx→π2−secxtanx

which is again in indeterminate form and if we try to apply the same we would get the function with which we have started, so l’Hospital’s Rule is not applicable in this case and we try to apply another method

limx→π2−secxtanx=limx→π2−1/cosxsinx/cosx=limx→π2−1sinx=1sinπ2=1

Final statement: The value of the limit is limx→π2−secxtanx=1