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We know that one of the characterizations of the exponential function is:

$$e^x=\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}$$

Trivially, it follows that $\lim_{n\rightarrow\infty}\left(1-\frac{x}{n}\right)^{n}=e^{-x}$

I am wondering about

$$\lim_{n\rightarrow\infty}\left(1-\frac{x}{n^{1+a}}\right)^{n}$$

where $a$ is a real number. Is the following evaluation of the expression correct?

$$\begin{array}{rcl}\lim_{n\rightarrow\infty}\left(1-\frac{x}{n^{1+a}}\right)^{n}&=&\lim_{n\rightarrow\infty}\left(1-\frac{xn^{-a}}{n}\right)^{n}\\ &=&\lim_{n\rightarrow\infty}e^{-xn^{-a}}\\ &=&\left\{\begin{array}{rl}1,&a>0\\e^{-x},&a=0\\0,&a<0\end{array}\right. \end{array}$$

I am uncomfortable taking the second equality, not sure what the justification is...

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  • $\begingroup$ Set $n^{1+a}=t \leftrightarrow t=n^{\frac{1}{1+a}}$ $\endgroup$ – Alex Oct 9 '12 at 1:59
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That step you are uncomfortable with is indeed quite dodgy. Instead, consider the logarithm of your limit. Since $\log(1-t) = -t+t^2/2 + t^3/3 + \cdots ,$

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

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

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  • $\begingroup$ Thanks! I knew that there was something wrong with that step... Your explanation makes sense. $\endgroup$ – M.B.M. Oct 9 '12 at 3:16

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