Question Number 103255 by Study last updated on 13/Jul/20

lim_(x→0) (((sin^6 (√2)x)/(tan^7 (x/2))))^(cos^(12) x) =???

Commented byDwaipayan Shikari last updated on 13/Jul/20

May be infinity

Answered by OlafThorendsen last updated on 13/Jul/20

sinu ∼_0 u tanu ∼_0 u Then ((sin^6 (√2)x)/(tan^7 (x/2))) ∼_0 ((((√2)x)^6 )/(((x/2))^7 )) = ((2^3 x^6 )/(x^7 /2^7 )) = ((1024)/x) And cosx ∼_0 1−(x^2 /2) ⇒ cos^(12) x ∼_0 1−6x^2 lim_(x→0) (((sin^6 (√2)x)/(tan^7 (x/2))))^(cos^(12) x) = lim_(x→0) (((1024)/x))^(1−6x^2 ) lim_(x→0) (((sin^6 (√2)x)/(tan^7 (x/2))))^(cos^(12) x) = { ((−∞ if x<0)),((+∞ if x>0)) :}

Commented byStudy last updated on 14/Jul/20

why cosx∼1−(x^2 /2)⇒cos^(12) x∼1−6x^2

Commented byOlafThorendsen last updated on 14/Jul/20

cosx ∼ 1−(x^2 /2) (1+u)^α ∼ 1+αu α = 12 and u = −(x^2 /2) Then cos^(12) x ∼ (1−(x^2 /2))^(12) ∼ 1−6x^2