# Limit of $\left(\frac{n^{2}+8n-1}{n^{2}-4n-5}\right)^{n}$, is the following true?

Using the previous question*, arithmetics of bordes, and the sandwich theorem, find the limits of the following sequenes:
b. $\left(\dfrac{n^{2}+8n-1}{n^{2}-4n-5}\right)^{n}$

(*) the previous question is to prove that $\lim_{n\to \infty }\left(1+\dfrac{x}{n}\right) = e^x$

Looking at the question it looked quite easy. Since we proved in class that $\lim (a_n)^k = (\lim a_n)^k$ I can easily say the limit of the inside is $1$, therefore the limit of everything is $1^n=1$ (This is a good time to notice that I don't know how to use anything but inline equations here):

$$\left(\frac{n^{2}+8n-1}{n^{2}-4n-5}\right)^{n}=\left(\dfrac{1+\dfrac{8}{n}-\dfrac{1}{n^{2}}}{1-\dfrac{4}{n}-\dfrac{5}{n^{2}}}\right)^{n}$$ due to arithmetics of borders:

$$\begin{align*} \lim\left(\frac{1+\dfrac{8}{n}-\dfrac{1}{n^{2}}}{1-\dfrac{4}{n}-\dfrac{5}{n^{2}}}\right)^{n}&=\left(\frac{\lim\left(1+\dfrac{8}{n}-\dfrac{1}{n^{2}}\right)}{\lim\left(1-\dfrac{4}{n}-\dfrac{5}{n^{2}}\right)}\right)^{n}\\\\ &=\left(\frac{\lim1+\lim\dfrac{n}{8}-\lim\dfrac{1}{n^{2}}}{\lim1-\lim\dfrac{4}{n}-\lim\dfrac{5}{n^{2}}}\right)^{n}\\\\ &=1^{n}\\\\ &=1 \end{align*}$$

But for some reason this feels wrong to me. It doesn't use the previous question or the sandwich theorem, and the solution feels to trivial to be true. Is there anything wrong here?

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With this logic, $e = \lim_{n \to \infty} (1+\frac{1}{n})^n=1^n=1$ – The Substitute Dec 9 '12 at 8:49
lol yeah... that's right... So I guess it's wrong that $\lim \left(\frac{n^{2}+8n-1}{n^{2}-4n-5}\right)^{n} = (\lim \left(\frac{n^{2}+8n-1}{n^{2}-4n-5}\right))^n$ huh?... Oh after thinking about this for a while (using the answer given after I started typing as well) I think I see why it's wrong. Thanks! – Nescio Dec 9 '12 at 8:58
The problem is that you can't take the $n$ outside of the limit. It is a dummy variable - it only has meaning in the context of a limit. Similarly, in calculus you can't write $\frac{d}{dx} x = x \frac{d}{dx} 1 = 0.$ – Jair Taylor Dec 9 '12 at 9:01

## 3 Answers

$$\lim_{n\to\infty}\left(\frac{n^{2}+8n-1}{n^{2}-4n-5}\right)^{n}$$ $$=\lim_{n\to\infty}\left(1+\frac{12n+4}{n^{2}-4n-5}\right)^{n}$$ $$=\lim_{n\to\infty}\left\{\left(1+\frac{12n+4}{n^{2}-4n-5}\right)^{\frac{n^{2}-4n-5}{12n+4}}\right\}^{\frac{(12n+4)n}{n^{2}-4n-5}}$$ $$=e^{12}$$ as $\lim_{n\to\infty}\frac{(12n+4)n}{n^{2}-4n-5}=\lim_{n\to\infty}\frac{12+\frac4n}{1-\frac4n-\frac5{n^2}}=12$

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Note that for any $\epsilon > 0$, $$1 + \dfrac{8-\epsilon}{n} \le 1 + \dfrac{8}{n} - \dfrac{1}{n^2} \le 1 + \dfrac{8}{n}$$ for sufficiently large $n$. Therefore $$\eqalign{e^{8-\epsilon} &= \lim_{n \to \infty} \left(1 + \dfrac{8-\epsilon}{n}\right)^n \le \liminf_{n \to \infty} \left(1 + \dfrac{8}{n} - \dfrac{1}{n^2}\right)^n \cr&\le \limsup_{n \to \infty} \left(1 + \dfrac{8}{n} - \dfrac{1}{n^2}\right)^n \le \lim_{n \to \infty} \left(1 + \dfrac{8}{n}\right)^n = e^8}$$ and taking $\epsilon \to 0+$ we get $$\lim_{n \to \infty} \left(1 + \dfrac{8}{n} - \dfrac{1}{n^2} \right)^n = \lim_{n \to \infty} \left(1 + \frac{8}{n} \right)^n = e^8$$ Similarly, $$\lim_{n \to \infty} \left(1 - \frac{4}{n} - \frac{5}{n^2} \right)^n = e^{-4}$$ So $$\lim_{n \to \infty} \left(\frac{n^2 - 8 n - 1}{n^2 - 4 n - 5}\right)^n = \dfrac{\displaystyle \lim_{n \to \infty} \left(1 + \dfrac{8}{n} - \dfrac{1}{n^2} \right)^n} {\displaystyle \lim_{n \to \infty} \left(1 - \dfrac{4}{n} - \dfrac{5}{n^2} \right)^n} = e^{12}$$

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Write \begin{equation}\left(\frac{n^{2}+8n-1}{n^{2}-4n-5}\right)^{n}=e^{\ln\left(\frac{n^{2}+8n-1}{n^{2}-4n-5}\right)n} \end{equation} and \begin{equation}\ln\left(\frac{n^{2}+8n-1}{n^{2}-4n-5}\right)n=\frac{\ln\left(\frac{n^{2}+8n-1}{n^{2}-4n-5}\right)}{\frac{1}{n}} \end{equation} FInally use De L'Hopital Rule to get $e^{12}$

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