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How do I begin to evaluate this limit: $\lim_{n\to \infty} \left(1-\frac{1}{n^2-4}\right)^{3n^2+5}$

Does it equal to $e^{-1}$? (Please don't use ln.)

Thanks a lot.

2 Answers 2

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Let $m=n^2-4$, and

$ \lim_{m\rightarrow\infty}\left(1-\frac1m\right)^{3m+17}=\left[\lim_{m\rightarrow\infty}\left(1-\frac1m\right)^{m}\right]^3=\frac1{e^3}$

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    Intuitively $m\rightarrow\infty$ such that 17 is really negligible in comparsion to it. If you are unhappy with this explanation, notice that $\lim_{m\rightarrow\infty}(1-\frac1m)^{17}=1$. As for the latter question, refer to the equation $m=n^2-4$.2012-05-02
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Let $v\longrightarrow‎ \infty$ and $‎u\longrightarrow‎ 1$ , then $\lim _{n‎\rightarrow \infty}u^v‎\longrightarrow‎ \exp (\lim_{n‎\rightarrow \infty} v(u-1))$.

So, we have $\lim_{n\to \infty} \left(1-\frac{1}{n^2-4}\right)^{3n^2+5}=exp(\lim_{n\to \infty} \left(1-\frac{1}{n^2-4}-1\right)({3n^2+5})=exp(\lim_{n\to \infty} \left(-\frac{3n^2+5}{n^2-4}\right))\\ =exp(-3)$