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I have the following limit to find:

$$\lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg ( \dfrac{2x^n}{x^n+1} \bigg)$$

Where $n \in \mathbb{N}^*$ and $x \in (0, \infty)$.

I almost got it. For $x > 1$, I observed that:

$$\lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg ( \dfrac{2x^n}{x^n+1} \bigg) = \lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg ( \dfrac{2x^n}{x^n(1 + \frac{1}{x^n})} \bigg) = \lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg ( \dfrac{2}{1+\frac{1}{x^n}} \bigg)$$

Because $x>1$, we have that $x^n \rightarrow \infty$ as $n \rightarrow \infty$, so that means that we have:

$$\dfrac{1}{\infty} \cdot \ln \bigg ( \dfrac{2}{1+\frac{1}{\infty}} \bigg ) = 0 \cdot \ln 2 = 0$$

The problem I have is in calculating for $x \in (0, 1]$. If we have that $x \in (0, 1]$ that means $x^n \rightarrow 0$ as $n \to \infty$, so:

$$\lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg( \dfrac{2x^n}{x^n + 1} \bigg ) = \lim\limits_{n \to \infty} \dfrac{\ln \bigg( \dfrac{2x^n}{x^n + 1}\bigg )}{n} $$

And I tried using L'Hospital and after a lot of calculation I ended up with

$$\ln x \lim\limits_{n \to \infty} \dfrac{x^n + 1}{x^n}$$

which is

$$\ln x\cdot \dfrac{1}{0}$$

And this is my problem. Maybe I applied L'Hospital incorrectly or something, I'm not sure. Long story short, I do not know how to calculate the following limit:

$$\lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg( \dfrac{2x^n}{x^n+1} \bigg )$$

when $x \in (0, 1]$.

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  • $\begingroup$ In the second part, $x\in (0,1)$ open. $\endgroup$ – user376343 Dec 25 '19 at 21:26
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No L'hopital needed - you just have to use the fact that $\ln(xy) = \ln(x) + \ln(y)$ and break up the limits.

$\lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg( \dfrac{2x^n}{x^n + 1} \bigg ) = $

$\lim\limits_{n \to \infty} \dfrac{\ln (2) + \ln(x^n) - \ln(x^n + 1)}{n} = $

$\lim\limits_{n \to \infty} \dfrac{\ln (2)}{n} + \lim\limits_{n \to \infty} \dfrac{n\cdot \ln(x)}{n} - \lim\limits_{n \to \infty} \dfrac{\ln(x^n + 1)}{n} = $

$ 0 + \ln(x)+ \lim\limits_{n \to \infty}\dfrac{\ln(x^n + 1)}{n} = \ln(x) $

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The term reduces to \begin{align*} \dfrac{1}{n}\log\left(2-\dfrac{2}{x^{n}+1}\right)&=\dfrac{1}{n}\log 2+\dfrac{1}{n}\log\left(1-\dfrac{1}{x^{n}+1}\right). \end{align*} We do the L'Hopital to the second term, it becomes \begin{align*} &\lim_{n\rightarrow\infty}\dfrac{\dfrac{1}{1-\dfrac{1}{x^{n}+1}}\dfrac{1}{(x^{n}+1)^{2}}x^{n}\log x}{1}\\ &=\lim_{n\rightarrow\infty}\dfrac{x^{n}+1}{x^{n}}\dfrac{1}{(x^{n}+1)^{2}}x^{n}\log x\\ &=\lim_{n\rightarrow\infty}(x^{n}+1)^{-1}\log x\\ &=\log x. \end{align*}

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Hint:

$(1/n)\log 2 +(1/n)\log x^n-(1/n)\log (x^n+1)=$

$(1/n)\log 2 + \log x -$

$(1/n)\log (x^n+1)$.

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Since $\frac{d}{dn}x^n=x^n\ln x$, $\frac{d}{dn}\frac{2x^n}{x^n+1}=-2\frac{d}{dn}\frac{1}{x^n+1}=\frac{2x^n\ln x}{(x^n+1)^2}$ and $\frac{d}{dn}\ln\frac{2x^n}{x^n+1}=\frac{\ln x}{x^n+1}$. So you want $\lim_{n\to\infty}\frac{\ln x}{x^n+1}$, regardless of $x$.

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You can use Cesàro theorem to obtain $$\lim_{n\to\infty} \frac{\ln 2 +n\ln x-\ln(x^{n}+1)}n = \lim_{n\to\infty}\left(\ln x-\ln\frac{x^{n+1}+1}{x^{n}+1}\right)\xrightarrow{n\to\infty} \ln x$$

since $\frac{x^{n+1}+1}{x^{n}+1} \to 1$ for $x \in (0,1]$.

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