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as in the topic I find it difficult to calculate following limit:$$\lim_{n\rightarrow\infty}(\frac{n-1}{n+1})^{b_{n}}$$where $b_{n}=(\sqrt{n+1}-\sqrt{n-1})^{-2}$. Any ideas and hints would be appreciated; thanks in advance!

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2 Answers 2

up vote 1 down vote accepted

This limit is of $1^\infty$. The standard way to solve them is by taking logarithms.

$P_n = (\frac{n-1}{n+1})^{b_n}$

Let $p_n = \log(P_n) = b_n \log(\frac{n-1}{n+1})$

By the standard expansion of log $$\log(\frac{n-1}{n+1}) = \log(1-\frac{2}{n+1}) \approx -\frac{2}{n+1} + O(\frac{1}{n^2})$$

By rationalization, $b_n$ can be written as

$$b_n = \frac{1}{(\sqrt{n+1}-\sqrt{n-1})^2} = \frac{(\sqrt{n+1}+\sqrt{n-1})^2}{({(n+1)}-{(n-1)})^2} = \frac{2(n+\sqrt{n^2-1})}{4}$$

So,$$p_n = \frac{2(n+\sqrt{n^2-1})}{4} (-\frac{2}{n+1}+O(n^{-2}))$$

$$p_n = -\frac{n+\sqrt{n^2-1}}{n+1} + O(n^{-1})$$

Taking the limit

$$\lim_{n \to \infty} p_n = -2$$

So,

$$\lim_{n \to \infty} P_n = \exp(-2)$$

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Hint: Rewrite this as $\left(1-\frac{a_n}n\right)^{nb_n}$ for some $(a_n)$ and $(b_n)$ such that $a_n\to a$ and $b_n\to b$. Then think about the behaviour of $\left(1-\frac{a}n\right)^{nb}$ when $n\to\infty$.

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