How can I calculate the limit $\lim\limits_{n \to \infty} \frac1n\ln \left( \frac{2x^n}{x^n+1} \right)$. 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 $x \in (0, 1]$.
 A: 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) $
A: 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*}
A: Hint:
$(1/n)\log 2 +(1/n)\log x^n-(1/n)\log (x^n+1)=$
$(1/n)\log 2 + \log x -$
$(1/n)\log (x^n+1)$.
A: 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 $x$.
A: 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 $x \in (0,1]$.
