http://www.slader.com/textbook/9781439049273-stewart-calculus-6th-edition/721/exercises/39/ PLease read problem in the link provided.
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lim n--> ∞ (1 + 2/n)^n
let 2/n = 1/m
n = 2m when n -->∞ m also -->∞
=> lim m--> ∞ (1 + 1/m)^2m
=> lim m--> ∞ [(1 + 1/m)^m ]^2
Recall that lim m--> ∞ (1 + 1/m)^m = e
Hence lim m--> ∞ [(1 + 1/m)^m ]^2 = e^2
let 2/n = 1/m
n = 2m when n -->∞ m also -->∞
=> lim m--> ∞ (1 + 1/m)^2m
=> lim m--> ∞ [(1 + 1/m)^m ]^2
Recall that lim m--> ∞ (1 + 1/m)^m = e
Hence lim m--> ∞ [(1 + 1/m)^m ]^2 = e^2
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use the limit definition of e to solve this.
e=lim as x approaches infinity of (1+1/x)^x
so substitute (2/n) for v and if v=2/n then n equals 2/v
so then you have lim as v approaches infinity for (1+v)^(2/v)
which= the lim as v approaches infinity for ((1+v)^(1/v))^2 and since the lim as v approaches infinity for "(1+v)^(1/v)" equals e and its being raised to the power of 2 you get e^2
you can also take the natural log of the whole thing, say the lne=1 and then bring 2 back as a power of e
e=lim as x approaches infinity of (1+1/x)^x
so substitute (2/n) for v and if v=2/n then n equals 2/v
so then you have lim as v approaches infinity for (1+v)^(2/v)
which= the lim as v approaches infinity for ((1+v)^(1/v))^2 and since the lim as v approaches infinity for "(1+v)^(1/v)" equals e and its being raised to the power of 2 you get e^2
you can also take the natural log of the whole thing, say the lne=1 and then bring 2 back as a power of e