What is the limit? lim x --> (pi/2), sec(x) - tan(x)

please show your steps!
and we are not allowed to use Hospital's rules

lim x --> (pi/2)( sec(x) - tan(x))

lim x-->pi/2 ((1-sin x)/cos x) = lim x-->pi/2 ((1-sin x)(cos x))/ cos^2 x

=lim x-->pi/2 ((1-sin x) cos x)/( 1-sin ^2 x) = lim x-->pi/2 (1-sin x)(cos x)/((1-sin x)(1+sin x))

=lim x-->pi/2 ((cos x)/(1+sin x)) = cos (pi/2)/((1+ sin(pi/2)) = 0/(1+1) = 0

If we convert sec(x) and tan(x) into terms of sines and cosines, we have:
lim (x-->π/2) [sec(x) - tan(x)] = lim (x-->π/2) [1/cos(x) - sin(x)/cos(x)]
= lim (x-->π/2) [1 - sin(x)]/cos(x).

Note that this limit takes the form 0/0, so we have to do some further manipulation.

Multiplying the numerator and denominator by 1 + sin(x):
lim (x-->π/2) [1 - sin(x)]/cos(x)
= lim (x-->π/2) {[1 + sin(x)][1 - sin(x)]}/{cos(x)[1 + sin(x)]}
= lim (x-->π/2) [1 - sin^2(x)]/{cos(x)[1 + sin(x)]}, via difference of squares
= lim (x-->π/2) cos^2(x)/{cos(x)[1 + sin(x)]}, since cos^2(x) = 1 - sin^2(x)
= lim (x-->π/2) cos(x)/[1 + sin(x)], by canceling cos(x)
= 0/(1 + 1), by evaluating the result at x = π/2
= 0.

I hope this helps!

are you suppose the use series or something?


1/cosx - sinx/cosx

(1-sinx)/cosx

limit as x approaches pi/2=

(1-0)/(0)=1/0=infinity. YOu don't even need L'Hospital's rule to do this problem.

If you don't believe me plug in 1/0.00000000001 into your calculator and you will get a very large number.

Here are two more ways to do this ...
——————————————————————————————————————
➊
   lim sec(x) - tan(x)
   x→π/2

= lim (1/cosx) - (sinx/cosx)
   x→π/2

= lim (1-sinx)/cosx
   x→π/2

= lim [cosx(1-sinx)] ̸ cos²x
   x→π/2

= lim [cosx(1-sinx)] ̸ (1-sin²x)
   x→π/2

= lim [cosx(1-sinx)] ̸ [(1+sinx)(1-sinx)]