1. ## Limit of e

Ok. I know that the lim (n to inf) of the sequence [1+(1/n)]^n=e.
How would I find the lim (n to inf) of [1+(2/n)]^n using the above limit? I know the answer is e^2, but how would I show it?

2. Originally Posted by JimmyT
Ok. I know that the lim (n to inf) of the sequence [1+(1/n)]^n=e.
How would I find the lim (n to inf) of [1+(2/n)]^n using the above limit? I know the answer is e^2, but how would I show it?
Hi,
this one looks like a trick - but it works:

$\left(1+\frac{2}{n}\right)^n$ = $\left(1+\frac{1}{\frac{n}{2}}\right)^n$ =
$\left( \left(1+\frac{1}{\frac{n}{2}}\right)^\frac{n}{2}\r ight)^2$

Now substitute n/2 = y. Then you get:

$\left( \left(1+\frac{1}{y}\right)^y\right)^2$

$\lim_{y\rightarrow \infty}{\left(1+\frac{1}{y}\right)^y}=e$

Therefore $\left( \left(1+\frac{1}{\frac{n}{2}}\right)^\frac{n}{2}\r ight)^2=e^2$

EB

3. Originally Posted by JimmyT
Ok. I know that the lim (n to inf) of the sequence [1+(1/n)]^n=e.
How would I find the lim (n to inf) of [1+(2/n)]^n using the above limit? I know the answer is e^2, but how would I show it?
If you want full rigour you will have to fill in the detail yourself but:

Put m=n/2, then

f(n)=[1+(2/n)]^n=[1+m]^(2m) = {[1+m]^m}^2

so

lim(n->infty) [1+(2/n)]^n = lim(m->infty) {[1+m]^m}^2
...................................= {lim(m->infty) [1+m]^m}^2=e^2.

RonL

4. Along the same lines, how would you manipulate the lim (n to inf) of
(3n)^(1/(2n)) to get 1?

Jim

5. Originally Posted by CaptainBlack
If you want full rigour you will have to fill in the detail yourself but:
Actually what Earboth did was rigorous. He used the limit composition rule for sequences.

6. Originally Posted by JimmyT
Along the same lines, how would you manipulate the lim (n to inf) of
(3n)^(1/(2n)) to get 1?

Jim
I am not going to write infinity because all limits of sequences are taken long infinite.

You have,
$\lim (3n)^{\frac{1}{2n}}=\lim 3^{\frac{1}{2n}}n^{\frac{1}{2n}}=\lim \sqrt{3}^{1/n}\cdot \sqrt{n}^{1/n}$
We divide the problem into two limit problems,

1) $\lim \sqrt{3}^{1/n}=\sqrt{3}^0=1$ (sequence composition rule).

2) $1\leq \sqrt{n}^{1/n}\leq n^{1/n}$
Squeeze theorem, the limit,
$\lim n^{1/n}=1$ is well known.

Thus the answer is $1\cdot 1 =1$