Which approximation allows for the following?
$$ \left(1-\frac{1}{x}\right)^n \approx e^{-n/x} $$
Here both $x$ and $n$ are variables.
$\endgroup$ 24 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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.
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.
$\endgroup$ $\begingroup$$\lim_\limits{n\to\infty} (1-\frac{x}{n})^n= e^{-x}$
Now replace x with n/x
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