Question Number 102347 by mohammad17 last updated on 08/Jul/20

Commented bybobhans last updated on 08/Jul/20

ln(y) = lim_(n→∞)  (1/n)(1+cos (((nπ)/2)))  ln(y) = lim_(n→∞) ((1+cos (((nπ)/2)))/n) = 0  y = e^0  = 1

Commented bymohammad17 last updated on 08/Jul/20

thank you sir

Answered by PRITHWISH SEN 2 last updated on 08/Jul/20

let  lnA=lim_(n→+∞)  (1/n)ln (1+cos((nπ)/2))       = lim_(n→+∞)   (1/n){cos ((nπ)/2)−(1/2)(cos ((nπ)/2))^2 +....}       = 0  (as cos oscillates between finite numbers)  ∴ A= 1   i.e. lim_(n→+∞)  (1+cos((n𝛑)/2))^(1/n) = 1 please check

Commented byDwaipayan Shikari last updated on 08/Jul/20

Yes it s right sir

Commented byPRITHWISH SEN 2 last updated on 08/Jul/20

thank you

Answered by mathmax by abdo last updated on 08/Jul/20

A_n =(1+cos(((nπ)/2)))^(1/n)  ⇒ A_n =e^((ln(1+cos(((nπ)/2))))/n)  but lim_(n→+∞)  ((ln(1+cos(((nπ)/2))))/n)=0 ⇒  lim_(n→+∞)  A_n =1