# Product of Limits of Real Sequences (1 + x over n)^n and (1 - x over n)^n equals 1

## Theorem

Let $\sequence {a_n}$ be the sequence defined as:

$a_n = \paren {1 + \dfrac x n}^n$

Let $\sequence {b_n}$ be the sequence defined as:

$b_n = \paren {1 - \dfrac x n}^n$

Then the product of the limits of $\sequence {a_n}$ and $\sequence {b_n}$ equals $1$

## Proof

From Real Sequence $\paren {1 + \dfrac x n}^n$ is Convergent, $\sequence {a_n}$ is convergent.

Setting $x \to -x$, it follows that $\sequence {\paren {1 + \dfrac {\paren {-x} } n}^n} = \sequence {b_n}$ is also convergent.

Then:

 $\displaystyle \paren {1 - \dfrac x n}^n \paren {1 - \dfrac x n}^n$ $=$ $\displaystyle \paren {\paren {1 - \dfrac x n} \paren {1 - \dfrac x n} }^n$ $\displaystyle$ $=$ $\displaystyle \paren {1 - \dfrac {x^2} {n^2} }^n$ Difference of Two Squares $\displaystyle$ $=$ $\displaystyle \paren {\paren {1 - \dfrac {x^2} {n^2} }^{n^2} }^{1/n}$

From Real Sequence $\paren {1 + \dfrac x n}^n$ is Convergent, $\sequence {\paren {1 - \dfrac {x^2} {n^2} }^{n^2} }$ is convergent to a positive limit.

It follows that:

$\exists a, b \in \R_{>0}: a < \paren {1 - \dfrac {x^2} {n^2} }^{n^2} < b$

for sufficiently large $n$.

Hence:

$a^{1/n} < \paren {1 - \dfrac x n}^n \paren {1 - \dfrac x n}^n < b^{1/n}$

The result follows from Limit of Root of Positive Real Number and the Squeeze Theorem for Real Sequences.

$\blacksquare$