Limit_Problem Ex_RP

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Dec 19th 2009, 09:51 AM
gilyos

Limit_Problem Ex_RP

$lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n*(\frac{2n+1}{n+1})^n$
Dec 19th 2009, 10:09 AM
galactus

Note that:

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

$=\frac{2n+1}{2n}=1+\frac{1}{2n}$

$\lim_{n\to \infty}\left(\frac{2n+1}{2n}\right)^{n}=\lim_{n\to \infty}\left(1+\frac{1}{2n}\right)^{n}$

Notice the similarity to the famous 'e' limit?.

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

So, what is the limit of $\lim_{n\to \infty}\left(1+\frac{1}{2n}\right)^{n}$
Dec 19th 2009, 10:12 AM
Raoh

Hint :
$\lim_{n\to \infty }\ln(1+\frac{1}{2n})^n= ?$
Dec 19th 2009, 10:26 AM
gilyos

Quote:

Originally Posted by Raoh

Hint :
$\lim_{n\to \infty }\ln(1+\frac{1}{2n})^n= ?$

$e^{\frac{1}{2}}$

but, what about :

$\lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n$
Dec 19th 2009, 10:35 AM
Raoh

Quote:

Originally Posted by Raoh

Hint :
$\lim_{n\to \infty }\ln(1+\frac{1}{2n})^n=?$

$\lim_{n\to \infty }\ln(1+\frac{1}{2n})^n=\frac{1}{2}$
Dec 19th 2009, 10:38 AM
Raoh

Quote:

Originally Posted by Raoh

$\lim_{n\to \infty }\ln(1+\frac{1}{2n})^n=\frac{1}{2}$

therefore your limit must be $e^{\frac{1}{2}}$ .
(Happy)
Dec 19th 2009, 12:38 PM
gilyos

O.k but what about
$\lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =???$
Dec 19th 2009, 12:55 PM
Raoh

Quote:

Originally Posted by gilyos

O.k but what about
$\lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =0$

i believe (Happy)
$\lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =0$
Dec 19th 2009, 12:57 PM
drop10

$<br /> \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n = 0<br />$
Why?

Intuition:

$<br /> (\frac{1}{2} + \frac{1}{2n}) <br />$
is always less than 1, and then raise it to any power greater than one, and it will get smaller.
Dec 19th 2009, 03:17 PM
mr fantastic

Quote:

Originally Posted by gilyos

O.k but what about
$\lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =???$

Given the discussion in this thread, a bit of thought might have suggested writing $\left( \frac{1}{2} + \frac{1}{2n}\right)^n = \left( \frac{1}{2}\right)^n \left( 1 + \frac{1}{n}\right)^n$ from which the limit is obvious.

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