Question

How do I solve the lim of (secx - tanx) as x --> pi/2?

Answer 1

Change it to \[\lim _{x \rightarrow \frac{ \pi }{ 2 }}\frac{ 1 }{ \cos(x) }-\frac{ \sin(x) }{ \cos(x) }\]
combine terms. Indeterminate as 0/0 -> apply l'Hopital.

Answer 2

you could write sec x as 1/cos x, and tan x as sin x / cos x
so
\[ \sec x - \tan x = \frac{1- \sin x}{\cos x} \]
multiply top and bottom by 1+ sin x
what do you get ?

Answer 3

(1-sin^2 x ) / (cosx)(1+sinx)

Answer 4

then i have cosx / 1+sinx

Answer 5

now take the limit

Answer 6

the answer is 0?

Answer 7

yes it approaches 0/2 = 0

Answer 8

thank you so much

Answer 9

yw