Limit of secx-tanx

Question

Limx--->(pi/2) secx-tanx

 

How do i solve this without hopital rule?

Answer 1

(a) Use the definitions sec x = 1/cos x and tan x = sin x / cos x

(b) combine fractions

(c) multiply the numerator and denominator by the conjugate of the numerator

 

(1-sin x)(1+sin x) / (cos x)(1+sin x)

 

(1 - sin²x) / (cos x)(1+sin x)

 

cos²x/(cos x)(1+sin x)

 

Now cancel and get lim 0/2