= lim cosx ̸ (1+sinx)
x→π/2
= cos(π/2) ̸ (1+sin(π/2))
= 0 ̸ 2
= 0 ← ANSWER
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➋
lim sec(x) - tan(x)
x→π/2
(sec x + tan x)
lim (sec x - tan x) ———————–
x→π/2 (sec x + tan x)
sec²x - tan²x
lim ———————–———–
x→π/2 (sec x + sec x·sin x)
(tan²x + 1) - tan²x
lim ———————–———–
x→π/2 sec x(1 + sin x)
1
lim ———————–———–
x→π/2 sec x(1 + sin x)
cos x cos(π/2) 0
lim ———————–— = ———–——–— = ———–— = 0
x→π/2 1 + sin x 1 + sin(π/2) 1 + 1
Have a good one!!
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