# Yet another evaluating limit question: $\lim_{n\to \infty} \left(1-\frac{1}{n^2-4}\right)^{3n^2+5}$

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.

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## 2 Answers

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? – Anonymous May 2 '12 at 9:55
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. – wangdw May 2 '12 at 10:04
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? – Anonymous May 2 '12 at 10:09
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. – André Nicolas May 2 '12 at 10:21
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 $m=n^2-4$. – wangdw May 2 '12 at 10:22

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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