# Calculate the limit of the string xn=[(n+1)/n]^(2n+3) n tends to +infinite

hala718 | Certified Educator

xn = [(n+1)/n]^(2n + 3)  n --> +inf.

lim xn = lim [(n+1)/n]^(2n + 3)

=[ lim(n+1)/n]^lim(2n+3)

Now divide and multiply by n:

==> lim xn = [lim ((1+1/n)/1]^ lim(2n+3)

= [lim(1+1/n)]^lim(2n+3)

= [lim (1+ 1/n)]^n)^lim (2n+3)/n

= e^lim(2n+3)/n

Now multiply and divide by n:

==> lim xn = e^lim(2+3/n)/1

When n --> inf, ==> lim xn = e^2

==> lim xn = e^2

giorgiana1976 | Student

We know the property that the limit will go to the base of the power function, also to the exponent.

lim xn = lim [(n+1)/n]^(2n+3)

But, according to the rule:

lim [(n+1)/n]^(2n+3) = lim [(n+1)/n]^lim (2n+3)

We'll calculate the limit of the base:

lim [(n+1)/n] = lim (n/n + 1/n) = lim (1 + 1/n) = lim1 + lim (1/n) = 1 + 1/inf. = 1+0 = 1

Now, we'll calculate the limit of the exponent:

lim (2n+3) = lim 2n + lim 3 = 2*inf + 3 = inf + 3 = inf

The limit of the ratio will be:

lim [(n+1)/n]^(2n+3) = 1^inf. which is an  indetermination case

But we know that:

lim (1 + 1/xn)^xn = e, when xn -> inf.

We'll create the limit:

lim (1 + 1/n)^lim (2n+3) = lim [(1 + 1/n)^n]^(2n+3)/n =

= e^lim(2n+3)/n

We'll calculate the limit of the exponent:

lim(2n+3)/n = lim n(2 + 3/n)/n = lim 2 + lim (3/n) = 2 + 0 = 2

So, the limit of the string is:

lim xn = lim [(n+1)/n]^(2n+3) = e^2

neela | Student

To fin ltxn. Xn = [(n+1)/n}^(2n+3) as n --> inf.

Lt (n+1/n) (2n+3) = lt{ (1+1/n)^n } ^(2n/n+3/n) as n-->nf

=  (e) ^2 , as  Lt (1+1/n)^n = e and Lt(2n/n+3/3) = 2 as n--> inf.

=e^2