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I’m trying to find the limit of $a_n = \left(1-\frac{1}{n^2}\right)^n$ for $n \rightarrow \infty$.

It seems that the limit is $1$, since $a_n = 0.999…$ for large $n$. The presentation $a_n = \frac{(n^2-1)^n}{n^{2n}}$ and expanding was my first idea, but I couldn’t get the result from there. Any ideas?

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By Bernoulli’s inequality

$$ 1 – \frac{1}{n} \leq \left(1 – \frac{1}{n^2}\right)^n \leq 1.$$

$$\left(1-\frac1{n^2}\right)^n=\left(1-\frac1n\right)^n\left(1+\frac1n\right)^n\xrightarrow[n\to\infty]{}e^{-1}e=1$$

Since you asked for an answer not using the limit leading to $1/e$ how about this?

Using the binomial expansion of $\left( 1-\dfrac1{n^2} \right)^n$ you should be able to set up a series in $n$ and then take the limit. This series will start out $$1^n + n \cdot 1^{n-1} \cdot \dfrac{-1}{n^2} + \dfrac{n \cdot (n-1)}2 \cdot1^{n-2} \cdot \left( \dfrac{-1}{n^2}\right) ^2 \ldots$$ and from there you should be able to simplify and show that the limit goes to $1$.

$$\lim_{n\to\infty}\left(1-\frac{1}{n^2}\right)^n=\lim_{n\to\infty}\left(\left(1-\frac{1}{n^2}\right)^{n^{2}}\right)^{\frac{1}{n}}=\left(\frac{1}{e}\right)^{0}=1$$

because

$$\lim_{n\to\infty}\left(1-\frac{1}{n^2}\right)^{n^{2}}=\frac{1}{e}$$ and

$$\lim_{n\to\infty}{\frac{1}{n}}=0$$

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