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