Limit of e

**JimmyT** · October 23rd 2006, 11:42 AM

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?

**earboth** · October 23rd 2006, 12:46 PM

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

**CaptainBlack** · October 23rd 2006, 12:52 PM

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

**JimmyT** · October 23rd 2006, 04:25 PM

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

Jim

**ThePerfectHacker** · October 23rd 2006, 04:50 PM

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.

**ThePerfectHacker** · October 23rd 2006, 04:57 PM

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 $1\cdot 1 =1$

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