find lim x-->(pi/2)^(-) (tan(x))^(cosx) hint:y=(tanx)^(cosx) (take logarithms)?
回答 (2)
2018-11-21 8:26 pm
y = [tan(x)]^cos(x)
Ln(y) = Ln[tan(x)]^cos(x) → when x approaches (π/2)-
Ln(y) = Ln[tan(x)]^cos(π/2)- → you know that: cos(π/2)- ≈ 0-
Ln(y) = Ln[tan(x)]^(0)- → you know that: a^(0) = 1 ← whatever the value of a
Ln(y) = Ln[1]-
Ln(y) = 0-
Si if Ln(y) approaches 0-, it means that y approaches 1-.
Lim [tan(x)]^cos(x) = 1-
x → (π/2)-
Ln(y) = Ln[tan(x)]^cos(x) → when x approaches (π/2)-
Ln(y) = Ln[tan(x)]^cos(π/2)- → you know that: cos(π/2)- ≈ 0-
Ln(y) = Ln[tan(x)]^(0)- → you know that: a^(0) = 1 ← whatever the value of a
Ln(y) = Ln[1]-
Ln(y) = 0-
Si if Ln(y) approaches 0-, it means that y approaches 1-.
Lim [tan(x)]^cos(x) = 1-
x → (π/2)-
2018-11-21 2:25 am
YES : ln y = cos x ln(tan x) = ln ( tan x ) / sec x ---> ∞ / ∞....
[ sec²x / tan x ] / [ sec x tan x ] = sec x / tan² x = cos x / sin²x ---> 0
thus y ---> 1
[ sec²x / tan x ] / [ sec x tan x ] = sec x / tan² x = cos x / sin²x ---> 0
thus y ---> 1
收錄日期: 2021-04-24 07:29:37
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