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Consider the function $z(n) = (1-f(n))^{g(n)}$. For $f(n) = \frac 1n, g(n) = n$ we have that $\lim z(n) = e^{-1}$; more generally, when $f(n) = \frac cn$ for any constant $c$, we have $\lim z(n) = e^{-c}$. In each of these cases we note that the limit is equal to $e^{-f(n)g(n)}$.

Here's my question: Under what conditions (on $f,g$) can we claim that $(1-f(n))^{g(n)} \sim e^{-f(n)g(n)}$? Unless $f(n)g(n)$ is a constant, the RHS also depends on $n$, which is why I'm only asking about asymptotic equality. My original guess was that it is sufficient to have $f(n) \to 0$ and $g(n) \to \infty$ but I haven't been able to get anywhere with a proof.

My context: I have the quantity $( 1 - \frac 1n)^{cn \log n}$ and I'd like desperately for this to be asymptotically equal to $e^{-c\log n} = n^{-c}$.

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closed as too localized by Brendan McKay, Suvrit, Gerry Myerson, Andres Caicedo, fedja Aug 18 '12 at 19:14

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This is much too elementary for this forum, voting to close. –  Brendan McKay Aug 18 '12 at 3:48
    
Good question, wrong website. MathOverflow is for questions of interest to researchers. Try math.stackexchange.com. –  Gerry Myerson Aug 18 '12 at 9:25
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3 Answers 3

This is a standard calculus question. Let $z(n) = (1−f(n))^{g(n)}$. Then $\log z(n) = g(n) \log( 1 - f(n))$. Assuming $f(n)\rightarrow 0$, one has $\log z(n)/( f(n)g(n) ) \rightarrow -1$. For your particular problem, all works as hoped for and your limit is $n^{-c}$.

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An easyish way to see that the first comment is valid:

If f(n) tends to 0, then

$\ln (1-f(n))$ tends to $-f(n)$ (take the first term of the Taylor series.)

So

$\ln (1-f(n))/f(n)$ tends to $-1$

So

$g(n) \ln (1-f(n))$ tends to $- g(n) f(n)$

Then exponentiate and the result follows.

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You need some sort of control over the $g(n)$:

Consider $f(n)=1/n$

$g(n)=n^2$

Then $(1-f(n))^{g(n)} \over \exp(-f(n)g(n))$ does not go to 1 as $n \rightarrow \infty$ (goes to maybe $\exp(-1/2)$ instead)

Ultimately the original functions are nice enough to work though.

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