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Which approximation allows for the following?

$$ \left(1-\frac{1}{x}\right)^n \approx e^{-n/x} $$

Here both $x$ and $n$ are variables.

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  • $\begingroup$ “Napier's original (anti)logarithms”? ;-) IIRC, Napier used $x=10^7$ and $n$ was what he called the logarithm of $(1-1/x)^n$. $\endgroup$ – egreg Apr 15 '16 at 22:03
  • $\begingroup$ You need to state which of n or x is getting large. $\endgroup$ – marty cohen Apr 15 '16 at 22:29
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The usual result (I don't know of a particular name for it) is that $$\lim_{n \to \infty} \left( 1 + \dfrac{t}{n}\right)^n = e^t $$ This is sometimes taken to be the definition of the exponential function.

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The $n$ doesn't matter: If $(1-\frac1x)\approx e^{-1/x}$, then I raise both sides to the $n$th power to get your approximation.
The Taylor series (or MacLaurin series) of $e^y$ is $e^y\approx 1+y+\frac{y^2}2+...$ The linear approximation is $e^y\approx 1+y$, which works when $y$ is small.
Now put $y=-1/x$, and get $1-\frac1x\approx e^{-1/x}$ which works when $x$ is large.

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The binomial theorem gives the approximation $$ \left(1-\frac{1}{x}\right)^n \approx {1-\frac{n}{x}} $$ and the series expansion of $e^{-n/x}$ gives the approximation $$ e^{-n/x} \approx {1-\frac{n}{x}} $$ hence the result.

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$\lim_\limits{n\to\infty} (1-\frac{x}{n})^n= e^{-x}$

Now replace x with n/x

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