Show that lim n approaches infinity (1 + 1/2n)^n= square root of e

Need help with the proof of this please!

consider the quantity e^(ln f(n) ), where f(n) = (1+1/2n)^n

Then the quantity becomes e^( n ln(1+1/2n) ), or e^( ln(1+s/2) / s), where we define s = 1/n

We are looking for limit as s approaches 0 (so n approaches infinity).

Apply L'Hopital's rule to the quantity inside the brackets -> derivative of top becomes 1/2 / (1+s/2). The derivative of the bottom becomes 1. Therefore, by L'Hopital's rule, the quantity becomes 1/2.

The quantity in question is therefoer e^(1/2)

QED

Hope that helps.

The Taylor series expansion of ln x is
Ln(1+x)=x-x^2/2+x^3/3-....
put x=k/n
Ln(1+k/n)=k/n-(k/n)^2/2+.....
nln(1+k/n)=k-k^2/2n+.....
ln((1+k/n)^n)=k-k^2/2n+...
putting k=1/2 and letting n go to inf, gives the result

Let L = the limit
L = lim (1 + 1/(2n))^n
......n->infinity
lnL = ln(lim (1 + 1/(2n))^n)
............n->infinity
lnL = lim ln((1 + 1/(2n))^n)
........n->infinity
lnL = lim n ln(1 + 1/(2n))
........n->infinity
lnL = lim ln(1 + 1/(2n))/(1/n)...now use L'Hopital's rule
........n->infinity
lnL = lim (-2n^-2/(1 + 1/(2n))/(-n^-2)
........n->infinity
lnL = lim (-2/(1 + 1/(2n))
........n->infinity
lnL = -2
e^(lnL) = e^-2
L = e^-2