2010/017) Lim (n -> infinity) {(1 + 1/n)^n - (1 + 1/n)}^-n is

(a) 1; (b) 1/(e - 1); (c) 1 - e^-1; (d) 0;?

Ans:
take (1+1/n) out to get

(1+1/n)^-n((1+1/n)^(n-1)-1)^-n

as n->infinite (1+1/n)^-n = 1/e

((1+1/n)^(n-1)-1)^(-n) = (e-1)^(-n) = 0

so product = 0

hence d

2010/009 Evaluate : Lim(n-> infinity) {(1 + 1/2n)(1 + 3/2n)(1 + 5/2n)...(1+(2n - 1)/2n)}^(1/2n)?

this can put as sum of

((1+1/2n)(1+2/2n) .....* * (1+(2n)/2n))^(1/2n)/
(1+2/2n)(1+4/2n) ........(1+ 2n/2n)^(1/2n)

basically multiply it by the missing terms and then devide back

now numerator say
N = ((1+1/2n)(1+2/2n) .....* * (1+(2n)/2n))^(1/2n)
take log to base e

log N = 1/2n (log (1+2/n) + log (1+ 2/2n) + .... + 2)

as n-> infinite this tehds to

1/2 int (log x) for x from 1 to 2

integral of log x = x log x - x
definite integral = (2 ln 2 -1) - (-1) =2 ln 2
so N = 1/2e^(2 ln 2) = e^2

now you can evaluate the denominator as
ln d = 1/2n (ln (1+2/2n ) + ln (1+4/2n ) + ...+ ln (2))
as n->infinite this is int log x

integral of log x = x log x - x
definite integral = (2 ln 2 -1) - (-1) =2 ln 2
so D = e^(2 ln 2) = 2e^2
so N/D = value = 1/2