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Finding $$\lim_{n\rightarrow \infty }\left(1-\frac{1}{n^2}\right)^n$$

mathlove
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E.H.E
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  • use $\lim_{n\to \infty}(1+x/n)^n = e^x$ – abel Jun 04 '15 at 23:03
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    Hint: $\left((1-1/n^2)^{n^2}\right)^{1/n}$. – André Nicolas Jun 04 '15 at 23:03
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    Intuitively: We know $\lim_{n\rightarrow \infty }\left(1-\frac{1}{n}\right)^n = e.$ But if you have $n^2$ on the inside without an $n^2$ in the exponent, the inside part converges faster to $1$ and the inside part "wins". – Jair Taylor Jun 04 '15 at 23:20
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    If one does not know the limit definition of $e$, then we can use Bernoulli's Inequality to write $$1\ge \left(1-\frac1{n^2}\right)^n\ge 1-\frac1n$$whereupon application of the squeeze theorem gives the limit as $1$ – Mark Viola Dec 02 '15 at 21:42
  • Sure. I up voted the question – Mark Viola Dec 02 '15 at 21:51

3 Answers3

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$$\lim_{n\rightarrow \infty }(1-\frac{1}{n^2})^n=\lim_{n\rightarrow \infty }(1-\frac{1}{n})^n(1+\frac{1}{n})^n=\frac{1}{e}e=1$$

E.H.E
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$b_n = \sqrt[n]{\left(1-\frac{1}{n^2}\right)^{n^2}}$. Since $a_n = \left(1-\frac{1}{n^2}\right)^{n^2}\to e^{-1}, |a_n-e^{-1}| < \dfrac{e^{-1}}{2}, n \geq N\to \dfrac{e^{-1}}{2} < a_n < \dfrac{3e^{-1}}{2}\Rightarrow \left(\dfrac{1}{2e}\right)^{\frac{1}{n}} < b_n < \left(\dfrac{3}{2e}\right)^{\frac{1}{n}}, n \geq N$ and by squeeze theorem $b_n \to 1$.

DeepSea
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Let's assume we replace 1/n^2 with an arbitrarily large number. Hence, 1/n^2 is insignificant and eventually becomes 0. Then, we know that 1 to the power of any number is still 1. Hence, the limit is 1

rassa45
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  • $1/n$ is also insignificant, right? So, if your argument is correct, then $\displaystyle\lim_{n\to\infty}(1-1/n)^n=1$. However, it is not true. This post can be helpful. – Pedro Jun 04 '15 at 23:47