I want to correctly prove the following statement:
\begin{equation} \lim_{x \rightarrow \infty} \frac{1}{x \cdot \ln\left(\frac{x + a}{x - a}\right)} = \frac{1}{2a} \ , \ a \in \mathbb{R}^{\ast+} \end{equation}
This should atleast be true for $a \in \mathbb{R}^{\ast+}$, but I have no idea how to correctly prove it, as the logarithm rules don't seem to be of any help here. When using the sum-statement of the logarithm for the denominator we should get:
\begin{equation} x \cdot \left(\frac{2a}{(x-a)} - \frac{2a^2}{(x-a)^2} + \frac{8a^3}{3(x-a)^3} - \frac{4a^4}{(x-a)^4} + HOT\right) \end{equation}
While the first term resembles the statement wo are looking for, can we actually just say that the rest of the terms tends to zero "fast" enough? And what would happen if $a$ tends to $\infty$ as well?
Edit: changed the statement so that it is mathematically correct. The previous statement was:
\begin{equation} \lim_{x \rightarrow \infty} \frac{1}{\ln\left(\frac{x + a}{x - a}\right)} = \frac{x}{2a} \ , \ a \in \mathbb{R}^{\ast+} \end{equation}
The funtion approaches $\infty$ in a linear way with the factor $\frac{1}{2a}$.
Consider $$ f(x)=\frac{1}{\ln\left(\frac{x+a}{x-a} \right)} = \frac{1}{\ln\left(1+\frac{2}{\frac{x}{a}-1} \right)} $$ With the variable change $t=\frac{x}{a}$, we have $$ g(t) = \frac{1}{\ln\left(1+\frac{2}{t-1} \right)} $$ Now we're interested in the value of $\lim_{t \to \infty} g(t)/t$. Using Taylor expansion for $\ln(1+x)$, we get $$ \ln(1+s) = s - \frac{s^2}{2} + \frac{s^3}{3} + \ldots $$ we get $\lim_{s \to 0} \ln(1+s) \approx s$. Therefore $$ \lim_{t\to \infty} \ln \left(1+\frac{2}{t} \right) \approx \frac{2}{t} $$ Combining the previous points, we get $$ \frac{1}{\ln\left(1+\frac{2}{t-1} \right)} \to \frac{1}{\left(\frac{2}{t} \right)} $$ and $$ \lim_{t \to \infty} \frac{g(t)}{t} \approx \frac{1}{t\left(\frac{2}{t} \right)} = \frac{1}{2} $$ Sorry for the slightly hand-wavy points, but I think this approach should yield the result.