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

• In the second part, $x\in (0,1)$ open. – user376343 Dec 25 '19 at 21:26

No L'hopital needed - you just have to use the fact that $$\ln(xy) = \ln(x) + \ln(y)$$ and break up the limits.

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

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

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

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

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*}

Hint:

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

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

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

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 $$x$$.

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