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...

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.

$\displaystyle {\lim _{x \to \infty }}{\left( {1 + \tfrac{1}{x}} \right)^x} = e\;\;\& \;{\lim _{x \to \infty }}{\left( {1 - \tfrac{1}{x}} \right)^x} = \frac{1}{e}$

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

Let $\displaystyle a, L $ be real numbers.

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

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

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

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

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

$\displaystyle \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 $

$\displaystyle = \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 $\displaystyle a = \ln{L} $.

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

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

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

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

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

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

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

$\displaystyle = e * \frac{1}{e} = 1 $

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.

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

Dear mathguy25:

After a long pause I am re-reading your answer, and...to tell the truth, I still understand about -10% of it!

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.

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........

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

Given that the limit of the product is the product of the limit, and given that fact, we have,

$\displaystyle \lim_{x \rightarrow \infty} \left ( 1 + \frac{1}{x}\right )^x = e$

$\displaystyle \lim_{x \rightarrow \infty} \left ( 1 - \frac{1}{x}\right )^x = \frac{1}{e}$

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

Which as said by mathguy25, and implied by plato, is equal to 1