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I am calculating the value of $\lim_{n\to\infty}(1-\frac{1}{n^x})^n$, where $x>0$. Here is my idea: let $z=n^x$, then $z$ tends to infinity as $n$ tends to infinity. Then $$\lim_{n\to\infty}(1-\frac{1}{n^x})^n=\lim_{z\to\infty}(1-\frac{1}{z})^{\log_{x}{z}}=\lim_{z\to\infty}((1-\frac{1}{z})^z)^\frac{\log_{x}z}{z}=\lim_{z\to\infty}(e^{-1})^0=1$$.
I'm not sure if the second to last equality holds.

Thanks for leablood's comment, I made a mistake. If $z=n^x$, then $n=z^{1/x}$.

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    if $z = n^x$ than $n \ne log_x z$. $n = z^{\frac 1 x}$. – 2017-01-31
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    if x<1,(1-1/n^x)^(n^x)*(n^a) where a+x =1, you can make (1-1/n^x)^(n^x) close to 1/e for big n, then rising power to n^a you can it goes to zero – 2017-01-31
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    Yes, I made a mistake. Then the result will coincide with Joseph's answer, which depends on $x$. – 2017-01-31

5 Answers 5

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Your equalities are wrong.

What you should do instead is:

$$(1-\frac 1{n^x})^n=e^{n\log(1-1/n^x)}.$$

Then use the fact that

$$\log(1+x)\underset{x\to 0}{\sim}x.$$

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$$(1-n^{-x})^n=\sum\limits_{k=0}^n\binom{n}{k}(-n^{-x})^k=\sum\limits_{k=0}^n \frac{(1-\frac{1}{n})…(1-\frac{k-1}{n})}{k!}(-n^{1-x})^k\approx e^{-n^{1-x}}$$

Put in a value for $x$ and let be $n\to\infty$ . Check this for $01$ .

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The log of the limit is $\frac{log(1-n^{-x})}{\frac{1}{n}}$. So using l'hopital, we have $\frac{xn^{-(x+1)}}{1-n^{-x}} \cdot -n^2$ which becomes $\frac{-xn}{n^x-1}$, differentiating again makes it clear that the log of the limit approaches zero. Therefore the limit approaches $1$ in most cases.

Note that the log of the limit becomes $\frac{-1}{n^{x-1}}$, so if $x>1$, the limit tends to $0$. If $1>x$, the limit tends to $-\infty$, suggesting that the limit in question is $1$ at $x> $ or equal to $1$.

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    Calculation of derivative of $\log(1-n^{-x})$ is wrong. – 2017-01-31
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    corrected, i think – 2017-01-31
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when n is large enough ${((1-1/n^x)^{n^x}})^{n^{1-x}}\approx (1/e)^{n^{1-x}}$, you can see it converges to 0, 1/e and 1 for appropriate values of x

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Let's consider the function instead of the sequence, which gives more flexibility. Taking the logarithm and setting set $u=1/t$, the limit becomes $$ \lim_{t\to\infty}t\log\left(1-\frac{1}{t^x}\right)= \lim_{u\to0^+}\frac{\log(1-u^x)}{u} $$ With l'Hôpital, $$ \lim_{u\to0^+}\frac{-xu^{x-1}}{1-u^x}= \begin{cases} -\infty & \text{if $01$} \end{cases} $$ Therefore, taking the exponential, $$ \lim_{t\to\infty}\left(1-\frac{1}{t^x}\right)^t= \begin{cases} 0 & \text{if $01$} \end{cases} $$