Limit of a sequence...

Find the limit of the sequence whose terms are given by an=(e^(2n)+6n)^(1/n)

this limit is for the form ∞^0

thus lim n-> ∞ e^(2n)+6n)^(1/n) = e^ ( lim n->∞ ln[(e^(2n) + 6n)]/n ) where ln is natural logarithm

now numerator and denominator there tends to ∞

hence by L'hospital rule

we have e^ ( lim n->∞ ln[(e^(2n) + 6n)]/n ) = e^ (lim n-> ∞ 2( e^(2n) + 3) / (e^(2n) + 6n) )

now divide by e^(2n) in both numerator and denominator,

we have e^ ( lim n-> ∞ 2( 1 + 3/e^(2n) ) / ( 1 + 6n/e^(2n) )

now as n-> ∞, e^(2n) grows to ∞ faster than 3 or 6n

hence lim n-> ∞ 3/e^(2n) or 6n/e^(2n) = 0

=> ans = e^2

e².....a_n = e^{ [ ln ( e^2n + 6n ) ] / n}...apply L'H to the exponent