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Specifically, is it correct to approximate the expression $\left(1 - 2^{-t}\right)^{x \cdot 2^t}$ as $e^{-x}$? This answer from an older post suggests that for an expression of the form $(1 - 1/n)^m$, quote

If $m$ depends of $n$ or the expression is part of a bigger term, it must be considered as a whole.

and so

[...] the sequence converges to 1.

which would contradict the approximation $e^{-x}$ of the original expression.

aayad
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    If $x$ is constant, then yes (since $(1-2^{-t})^{x\cdot 2^t} = \bigl( (1-2^{-t})^{2^t} \bigr)^x$). In more complicated situations: always take logarithms, try to understand the limit of $m\log(1-1/n)$, and then exponentiate the answer. – Greg Martin Feb 01 '24 at 23:18
  • Then, if, say $t= cx$ for constant $c$ (this happens to be my specific case of interest), the expression $(1 - 2^{-t})^{x \cdot 2^t}$ works out to $e^{-t}$ by taking log and exponentiating, which is $0$ in the limit $t \to \infty$. – aayad Feb 01 '24 at 23:45

1 Answers1

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We have that for any $f(t) \to \infty$

$$\left(1-\frac1{f(t)}\right)^{x\cdot f(t)} \to e^{-x}$$

which is related and can be proved by the standard limit

$$\left(1+\frac1{n}\right)^{n} \to e$$

The case $(1 - 1/n)^m$ with $n\to \infty$ and $m\to \infty$ is undetermined if we don't have any information about the rate of growing of $m$ with respect to $n$, indeed for example

  • $m=kn$

$$\left(1+\frac1n\right)^{kn} \to e^k$$

  • $m=n^2$

$$\left(1+\frac1n\right)^{n^2} \to \infty$$

user
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