Question

Lt [(3n+1)/(3n-1)]^n =?
n->infty

Answer 1

lim_(n->infinity) ((3 n+1)/(3 n-1))^n
Indeterminate form of type 1^infinity. Transform using lim_(n->infinity) ((3 n+1)/(3 n-1))^n = e^(lim_(n->infinity) n log((3 n+1)/(3 n-1))):
= e^(lim_(n->infinity) n log((3 n+1)/(3 n-1)))
Indeterminate form of type 0·infinity. Let t = 1/n, then lim_(n->infinity) n log((3 n+1)/(3 n-1)) = lim_(t->0) (log((3/t+1)/(3/t-1)))/t:
= e^(lim_(t->0) (log((1+3/t)/(3/t-1)))/t)
Indeterminate form of type 0/0. Applying L'Hospital's rule we have, lim_(t->0) (log((1+3/t)/(-1+3/t)))/t = lim_(t->0) ( d/( dt) log((1+3/t)/(-1+3/t)))/(( dt)/( dt)):
= e^(lim_(t->0) -6/(t^2-9))
Factor out constants:
= 1/e^(6 (lim_(t->0) 1/(-9+t^2)))
The limit of a quotient is the quotient of the limits:
= 1/e^(6/(lim_(t->0) (-9+t^2)))
The limit of t^2-9 as t approaches 0 is -9:
= e^(2/3)

Answer 2

lim n-->infnity[ (1+x/n)^1/n ]=e^(x)
so taking 3n out from above and bottom of the expression
it'll become e^(1/3)/e^(-1/3)=e^(2/3)