Evaluating $\underset{n\to \mathrm{\infty}}{lim}{(1-\frac{x}{{n}^{1+a}})}^{n}$

Garrett Sheppard
2022-08-14
Answered

Evaluating $\underset{n\to \mathrm{\infty}}{lim}{(1-\frac{x}{{n}^{1+a}})}^{n}$

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Brennan Parks

Answered 2022-08-15
Author has **14** answers

That step you are uncomfortable with is indeed quite dodgy. Instead, consider the logarithm of your limit. Since $\mathrm{log}(1-t)=-t+{t}^{2}/2+{t}^{3}/3+\cdots ,$

$n\mathrm{log}(1-\frac{x}{{n}^{1+a}})=-(\frac{x}{{n}^{a}}+\frac{{x}^{2}}{2{n}^{1+2a}}+\frac{{x}^{3}}{3{n}^{2+3a}}+\cdots )$

so as $n\to \mathrm{\infty},$ if $a=0$ this goes to $-x$, if $a>0$ to goes to 0 and if $a<0$ this goes to $-\mathrm{\infty}.$. Exponentiating gives back the result.

$n\mathrm{log}(1-\frac{x}{{n}^{1+a}})=-(\frac{x}{{n}^{a}}+\frac{{x}^{2}}{2{n}^{1+2a}}+\frac{{x}^{3}}{3{n}^{2+3a}}+\cdots )$

so as $n\to \mathrm{\infty},$ if $a=0$ this goes to $-x$, if $a>0$ to goes to 0 and if $a<0$ this goes to $-\mathrm{\infty}.$. Exponentiating gives back the result.

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