# limit proof

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• Dec 13th 2010, 03:46 AM
HallsofIvy
Quote:

Originally Posted by alexandros
Can research show us that:

$\displaystyle \sqrt 2 = \frac{3}{2}$??

$\displaystyle \left(1+\frac{1}{n}\right)^{nh}= 1+h+\frac{(nh)(nh-1)}{2!}\;\frac{1}{n^2}+.....$
for n= 1 and h= 1/2?

The left side is equal to $\displaystyle \sqrt{2}$. The right hand side is certainly NOT "3/2".
The first four partial sums are 1, 1.5, 1.375, and 1.4375. Since that is an alternating sum with decreasing terms, its sum lies between any two consecutive partial sums. In particular, the sum lies between 1.375 and 1.4375. $\displaystyle \sqrt{2}$, approximately 1.414, lies between those two, 3/2= 1.5 does not.
• Dec 13th 2010, 04:10 AM
Quote:

Originally Posted by alexandros
Can research show us that:

$\displaystyle \sqrt 2 = \frac{3}{2}$??

What I wrote was "truncated"...

I thought that would be easily deduced.

Maybe an easier solution to follow would be...

$\displaystyle \displaystyle\lim_{h \to 0}\frac{e^h-1}{h}=?$

Substitute $\displaystyle \displaystyle\ e^h-1=\frac{1}{t}\Rightarrow\ e^h=1+\frac{1}{t}$

$\displaystyle \displaystyle\Rightarrow\ h=ln\left[1+\frac{1}{t}\right]$

$\displaystyle h\rightarrow\ 0+\ \Rightarrow\ t\rightarrow \infty$

$\displaystyle \displaystyle\Rightarrow\lim_{t \to \infty}\frac{1}{tln\left[1+\frac{1}{t}\right]}$

$\displaystyle =\displaystyle\lim_{t \to \infty}\frac{1}{ln\left(1+\frac{1}{t}\right)^t}=\f rac{1}{ln(e)}=1$
• Dec 21st 2010, 10:24 AM
alexandros
Quote:

Originally Posted by HallsofIvy
$\displaystyle \left(1+\frac{1}{n}\right)^{nh}= 1+h+\frac{(nh)(nh-1)}{2!}\;\frac{1}{n^2}+.....$
The left side is equal to $\displaystyle \sqrt{2}$. The right hand side is certainly NOT "3/2".
The first four partial sums are 1, 1.5, 1.375, and 1.4375. Since that is an alternating sum with decreasing terms, its sum lies between any two consecutive partial sums. In particular, the sum lies between 1.375 and 1.4375. $\displaystyle \sqrt{2}$, approximately 1.414, lies between those two, 3/2= 1.5 does not.