# Limit_Problem Ex_RP

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

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

$\lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n$
• 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
$\lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =???$
• Dec 19th 2009, 12:55 PM
Raoh
Quote:

Originally Posted by gilyos
$\lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =0$

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

Why?

Intuition:

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

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