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 $x \in (0, 1]$.

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) $
