Undergrad Significance of (1+1/x)^x as x->-inf

[feynman1]

Any practical or scientific significance of (1+1/x)^x as x->-inf?

 

[anuttarasammyak]

$e$

 

[pasmith]

Any practical or scientific significance of (1+1/x)^x as x->-inf?

If interest of r APR is compounded monthly, then after t years the balance of the account will be \left(1 + \frac{r}{12}\right)^{12t}. Now imagine that interest is instead compounded every 1/nth of a year. Then after t years the balance is \left(1 + \frac{r}{n}\right)^{nt}. Now take the limit as n \to \infty. This is known as "continuous compounding" and after t years the balance of the account is e^{rt}.

 

[feynman1]

If interest of r APR is compounded monthly, then after t years the balance of the account will be \left(1 + \frac{r}{12}\right)^{12t}. Now imagine that interest is instead compounded every 1/nth of a year. Then after t years the balance is \left(1 + \frac{r}{n}\right)^{nt}. Now take the limit as n \to \infty. This is known as "continuous compounding" and after t years the balance of the account is e^{rt}.

x->-inf

 

[fresh_42]

The sign doesn't matter.

x->-inf

\begin{align*}
\left(1+\dfrac{1}{x}\right)^x&=\left(1-\dfrac{1}{|x|}\right)^{-|x|}
=\left(\dfrac{1}{1-\dfrac{1}{|x|}}\right)^{|x|}=\left(\dfrac{|x|}{|x|-1}\right)^{|x|}\\
&=\left(1+\dfrac{1}{|x|}+\dfrac{1}{|x|^2}+\ldots\right)^{|x|}\stackrel{|x|\to\infty }{\longrightarrow }\lim_{|x|\to\infty }\left(1+\dfrac{1}{|x|}\right)^{|x|}=e
\end{align*}

 

[feynman1]

The sign doesn't matter.

\begin{align*}
\left(1+\dfrac{1}{x}\right)^x&=\left(1-\dfrac{1}{|x|}\right)^{-|x|}
=\left(\dfrac{1}{1-\dfrac{1}{|x|}}\right)^{|x|}=\left(\dfrac{|x|}{|x|-1}\right)^{|x|}\\
&=\left(1+\dfrac{1}{|x|}+\dfrac{1}{|x|^2}+\ldots\right)^{|x|}\stackrel{|x|\to\infty }{\longrightarrow }\lim_{|x|\to\infty }\left(1+\dfrac{1}{|x|}\right)^{|x|}=e
\end{align*}

the derivation was known, but was asking about the practical meaning of -inf, not maths

 

[pasmith]

x->-inf

If you reduce the balance by r/n every 1/n years then the balance is <br /> \left(1 - \frac{r}{n}\right)^{n} = \left(1 + \frac rn\right)^{-n}.