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According to WolframAlpha, the limit of

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

and I wonder how this result is obtained.

My approach would be to divide both nominator and denominator by $n$, yielding

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

As $ \frac{1}{n} \to 0 $ as $ n \to \infty$, what remains is

$$ \lim_{n \to \infty} \left(\frac{1-0}{1 + 0} \right)^{2n+4} = 1 $$

What's wrong with my approach?

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    $\begingroup$ $\lim_{n \to \infty} (\dfrac{1}{n}+1)^{n}=e$ is a Classic Limit definition of e. $\endgroup$
    – Inceptio
    Jun 23, 2013 at 9:33
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    $\begingroup$ The wrongness in your approach is in assuming that you can freely replace $ \lim_{n \to \infty} \left(\frac{1}{n}\right)$ by 0 (exactly zero). Loosely speaking: $ 1/n $ becomes "nearly" zero, so $ 1-1/n $ becomes "nearly" one, yet that "very small" difference may be relevant after "infinitely" many multiplications. A reference case (the reference case) being the limit shown by Inceptio in another comment $\endgroup$
    – Rolazaro Azeveires
    Jun 23, 2013 at 13:25

3 Answers 3

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If you already know that

$$\lim_{n\to\infty}\left(1+\frac x{f(n)}\right)^{f(n)}=e^x\;,\;\;\forall\,x\in\Bbb R$$

and for any function $\,f\,$ s.t. $\,f(n)\xrightarrow[n\to\infty]{}\infty\;$ , then

$$\left(\frac{n-1}{n+1}\right)^{2n+4}=\left[\left(1-\frac2{n+1}\right)^{n+1}\right]^2\left(1-\frac2{n+1}\right)^2\xrightarrow[n\to\infty]{}(e^{-2})^2\cdot1^2=e^{-4}$$

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  • $\begingroup$ Alright, that makes sens! However, what's wrong with my approach then? Do the best of my knowledge, I made only valid operations, yet the result is different. $\endgroup$
    – bonifaz
    Jun 23, 2013 at 9:35
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    $\begingroup$ You tried to apply arithmetic of limits selectively and thus wrongly, i.e.: you made $\,n\to \infty\,$ within the parentheses (and thus you get in there, correctly, $\;1\;$) , yet you kept the exponent, which is also a function of $\,n\,$, constant at the same time! You can't do that...:) $\endgroup$
    – DonAntonio
    Jun 23, 2013 at 9:47
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The definition of $e$ is

$$\lim_{n \to \infty} \left (1+\frac{1}{n}\right)^n$$

There are many ways in which this may be shown. One is to use logarithms: let $L$ be the limit; then

$$\log{L} =\lim_{n \to \infty} n \log{\left (1+\frac{1}{n}\right)} = \lim_{n \to \infty} n \frac{1}{n} = 1$$

so that $L = e$. Further, you can show similarly that

$$\lim_{n \to \infty} \left (1+\frac{x}{n}\right)^n = e^x$$

Your limit is then

$$\left ( \frac{1/e}{e}\right)^2 = \frac{1}{e^4}$$

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

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

$$ = \left(\lim_{n \to \infty}\left(1+\frac{(-2)}{n+1} \right)^{\frac{n+1}{-2}}\right)^{\lim_{n \to \infty}\left(\frac{-4-\frac2n}{1+\frac1n}\right)}$$

$$=(e)^{-4}\text{ as } n \to \infty, \frac{n+1}{-2}\to-\infty\text{ and } \lim_{m\to\infty}\left(1+\frac1m\right)^m=e$$

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