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I would like to prove $\lim_{n\rightarrow\infty}(1-n^{-\alpha+o(1)})^n$. $0< \alpha<1$.

I asked my friend and he telled me that $$(1-n^{-\alpha+o(1)})^n=[(1-n^{-\alpha+o(1)})^{n^{\alpha+o(1)}})]^{n^{1-\alpha-o(1)}}$$.

Since $\lim_{n\rightarrow\infty}(1-n^{-\alpha+o(1)})^{n^{\alpha+o(1)}}=e^{-1}$, we have $$\lim_{n\rightarrow\infty}(1-n^{-\alpha+o(1)})^n= \lim_{n\rightarrow\infty}e^{-n^{1-\alpha-o(1)}}=0$$ I don't think it is rigorous. I have the following two main questions.

(1) How to show $\lim_{n\rightarrow\infty}(1-n^{-\alpha+o(1)})^{n^{\alpha+o(1)}}=e^{-1}$? I know $\lim_{n\rightarrow\infty}(1-n^{-\alpha})^{n^{\alpha}}=e^{-1}$ but I don't know how to deal with extra o(1).

(2) How can we somehow first evaluate the limit inside the exponential? By the same reasoning, I think we can have $\lim_{n\rightarrow\infty}(1-n^{-\alpha+o(1)})^n=\lim_{n\rightarrow\infty}(\lim_{n\rightarrow\infty}(1-n^{-\alpha+o(1)}))^n=\lim_{n\rightarrow\infty}1^n=1$, which results in a totally different answer.

Wei-Cheng Lee
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  • Here is one approach – Mittens Oct 23 '22 at 04:21
  • If we want to use that approach, we have to set $c_n=-n^{1-\alpha+o(1)}$, which does not converge. How could we use that approach? – Wei-Cheng Lee Oct 23 '22 at 07:44
  • It diverges to $-\infty$, provided $o(1)$ is real, which still works. In any case, for a real sequence $a_n\rightarrow 0$, $$E_n:=\big(1-\frac{1}{n^{\alpha+a_n}}\Big)^{n^{\alpha+a_n}}\xrightarrow{n\rightarrow\infty}e^{-1}$$ since $0<e^{-1}<1$, for all $n$ large enough, say all $n\geq N$ for some $N$, $E:=\sup_{n\geq N}E_n<1$. $E^{n^{1-\alpha-a_n}}\rightarrow0$ – Mittens Oct 23 '22 at 10:31
  • I see how the first approach works. However, I can't understand why in the proof we have $$max_{1\leq j \leq n}{|z_j|,|w_j|}\leq e ^{\frac{\gamma}{n}}$$ We have no control for $n< n_0$. Sorry to comment here, I don't have enough reputaion to comment there. – Wei-Cheng Lee Oct 26 '22 at 13:08

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