# Why ((1+1/x)^x) * ((1-1/x)^x) approaches 1 from "zero" rather than "2" side?

• May 2nd 2013, 04:58 AM
kpkkpk
Why ((1+1/x)^x) * ((1-1/x)^x) approaches 1 from "zero" rather than "2" side?
(1+1/x)^x approaches about 2,71... when x grows bigger. We mark this as e (=base number of natural logarithm).
I tried what happens with (1-1/x)^x: it seems to approach about 0,3676... as x grows bigger.
When I raise this 0,3676...to -1, it then seems to be = e.

When I then tried, what happens if I multiply (1+1/x)^x with ((1-1/x)^x, it was not a great surprise for me that they seemed to approach number 1...as x grows bigger and bigger.

But why does it so from one side only? Why does it not oscillate between both sides of number 1? ...or why does it not approach number 1(Wondering) from bigger side?

Below there are some numbers to clarify what I mean:
(I really hope that this program will not blurr my presentation by moving these numbers and characters in inappropriate places, but my apologize if it does anyway...I am not very gifted in this field!)

X (1+1/x)^x (1-1/x)^x (1+1/x)^x * (1-1/x)^x

0 1 ?
1 2 0 0
2 2,25 0,25 0,5625
3 2,37037037... 0,296... 0,702331961...
4 2,44140625... 0,31640625... 0,772476196...
5 2,48832 0,32768 0,815372697
6 2,521626372... 0,334897976... 0,844487569...
7 2,546499697... 0,339916677... 0,865597715...
8 2,56784514... 0,343608915... 0,881626435...
9 2,581174792... 0,346439416... 0,894220687...
10 2,59374246... 0,34867844... 0,904382075...
50 2,691588029... 0,36416968... 0,980194751...
100 2,704813829... 0,366032341... 0,990049338...
1000 2,716923932... 0,367695424... 0,999000499...
• May 2nd 2013, 05:34 AM
Plato
Re: Why ((1+1/x)^x) * ((1-1/x)^x) approaches 1 from "zero" rather than "2" side?
Quote:

Originally Posted by kpkkpk
(1+1/x)^x approaches about 2,71... when x grows bigger. We mark this as e (=base number of natural logarithm).
I tried what happens with (1-1/x)^x: it seems to approach about 0,3676... as x grows bigger.
When I raise this 0,3676...to -1, it then seems to be = e.

${\lim _{x \to \infty }}{\left( {1 + \tfrac{1}{x}} \right)^x} = e\;\;\& \;{\lim _{x \to \infty }}{\left( {1 - \tfrac{1}{x}} \right)^x} = \frac{1}{e}$
• May 2nd 2013, 07:18 AM
mathguy25
Re: Why ((1+1/x)^x) * ((1-1/x)^x) approaches 1 from "zero" rather than "2" side?
Let $a, L$ be real numbers.

Let $\lim_{x \rightarrow \infty} (1 + \frac{a}{x})^x = L$.

Then $\lim_{x \rightarrow \infty} \ln{(1 + \frac{a}{x})^x} = \ln{L}$

Then $\lim_{x \rightarrow \infty} x * \ln{(1 + \frac{a}{x})} = \ln{L}$

Then $\lim_{x \rightarrow \infty} \frac{\ln{(1+\frac{a}{x})}}{\frac{1}{x}} = \ln{L}$

The limit here is of the form $\frac{0}{0}$. Use L'hopital's rule!

$\lim_{x \rightarrow \infty} \frac{\ln{(1 + \frac{a}{x})}}{\frac{1}{x}} = \lim_{x \rightarrow \infty} \frac{\frac{1}{1+\frac{a}{x}} * \frac{-a}{x^2}}{\frac{-1}{x^2}} = \lim_{x \rightarrow \infty} \frac{1}{1 + \frac{a}{x}} * \frac{-a}{x^2} * -x^2$
$= \lim_{x \rightarrow \infty} \frac{a}{1 + \frac{a}{x}} = a \lim_{x \rightarrow \infty} \frac{1}{1 + \frac{a}{x}} = \frac{a}{1 + 0} = a$.

Then $a = \ln{L}$.

Then $e^a = e^\ln{L} = L$.

Therefore, $\lim_{x \rightarrow \infty} (1 + \frac{a}{x})^x = e^a$.

Then when $a = 1, - 1$ we see that

$\lim_{x \rightarrow \infty} (1 + \frac{1}{x})^x = e^1 = e$ and
$\lim_{x \rightarrow \infty} (1 + \frac{-1}{x})^x = \lim_{x \rightarrow \infty} (1 - \frac{1}{x})^x = e^{-1} = \frac{1}{e}$

Therefore, $\lim_{x \rightarrow \infty} (1 + \frac{1}{x})^x * (1 - \frac{1}{x})^x$

$= \lim_{x \rightarrow \infty} (1 + \frac{1}{x})^x * \lim_{x \rightarrow \infty} (1 - \frac{1}{x})^x$

$= e * \frac{1}{e} = 1$
• May 6th 2013, 12:58 AM
kpkkpk
Re: Why ((1+1/x)^x) * ((1-1/x)^x) approaches 1 from "zero" rather than "2" side?
Thank You both who answered to me!
As I was afraid numbers in my little "table" are difficult to read; sorry about that.
• June 3rd 2013, 03:24 AM
kpkkpk
Re: Why ((1+1/x)^x) * ((1-1/x)^x) approaches 1 from "zero" rather than "2" side?
Dear mathguy25:
• June 3rd 2013, 03:57 AM
Shakarri
Re: Why ((1+1/x)^x) * ((1-1/x)^x) approaches 1 from "zero" rather than "2" side?
The expression for e approaches from the lower side and the expression for 1/e approaches from the lower side so the expression for e*(1/e) would approach 1 from the lower side.
• June 3rd 2013, 05:43 AM
MINOANMAN
Re: Why ((1+1/x)^x) * ((1-1/x)^x) approaches 1 from "zero" rather than "2" side?
Kp...

study limits..and you wil understand all these........
• June 3rd 2013, 05:59 AM
$\lim_{x \rightarrow \infty} \left ( 1 + \frac{1}{x}\right )^x = e$
$\lim_{x \rightarrow \infty} \left ( 1 - \frac{1}{x}\right )^x = \frac{1}{e}$
$\lim_{x \rightarrow \infty} \left (\left ( 1 + \frac{1}{x}\right )^x \left ( 1 - \frac{1}{x}\right )^x \right ) = \lim_{x \rightarrow \infty} \left ( 1 + \frac{1}{x}\right )^x \lim_{x \rightarrow \infty} \left ( 1 - \frac{1}{x}\right )^x = e \frac{1}{e} = 1$