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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.

3 Answers 3

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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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    Thanks, is there a way to prove this without changing n to m?2012-05-02
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    The basic idea behind this proof is to transform the limit to a representation of the natural base $e$ (As you prohibit the use of $\ln$). So, mi dispiace I cannot think of a way without doing this $m$ thing.2012-05-02
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    OK then why is it ok to remove the 17 from the exponent? also, why doesn't it matter if m tends to infinity isntead of n?2012-05-02
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    Formally, $\left(1-\frac{1}{m}\right)^{3m+17}=\left(1-\frac{1}{m}\right)^{3m}\left(1-\frac{1}{m}\right)^{17}$. But $\left(1-\frac{1}{m}\right)^{17}$ approaches $1$ as $m\to\infty$, so does not affect the limit.2012-05-02
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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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You can also use the well known series expansion of $\exp(-x) = 1 - 1/x + O(x^{-2})$:

$$\begin{align} \lim_{n\to \infty} \left(1-\frac{1}{n^2-4}\right)^{3n^2+5} &= \lim_{n\to \infty} \exp\left(-\frac{3n^2+5}{n^2-4}\right) + O(n^{-2})\\ &= \lim_{n\to \infty} \exp\left(-3 + O(n^{-2})\right)\\ &= \exp(-3) \end{align}$$

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