Limit (without series expansion and l'Hôpital's rule) $$\lim_{x \to \infty}\ln{\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}}}\cdot \left(\ln{\frac{x+1}{x-1}}\right)^{-2}=\frac{1}{8}$$
Any suggestion to find this limit without series expansion and l'Hôpital's rule? Thanks and regards.
Note:
WolframAlpha confirms that the result is $\frac{1}{8}$.
 A: I guess the only way to avoid using series expansion or l'Hôpital's rule is to reduce the limit to known limits. Let $y=1/x$. Then
$$
   \lim_{x \to \infty}\ln{\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}}}\cdot \left(\ln{\frac{x+1}{x-1}}\right)^{-2} = \lim_{y \to 0} \frac{\ln \frac{1+\sqrt{1+y^2}}{1+\sqrt{1-y^2}}}{\left(\ln \frac{1+y}{1-y}\right)^2} = \lim_{y \to 0} \frac{\frac{1}{y^2}\ln\left(\frac{1+\sqrt{1+y^2}}{2}\right) - \frac{1}{y^2}\ln\left(\frac{1+\sqrt{1-y^2}}{2}\right)}{\left(\frac{1}{y} \ln(1+y) - \frac{1}{y} \ln(1-y)\right)^2}
$$
Now, using:
$$
   \lim_{y \to 0} \frac{1}{y} \ln(1\pm y) = \pm 1,  \qquad \lim_{y \to 0} \frac{1}{y^2} \ln\left(\frac{1+\sqrt{1 \pm y^2}}{2}\right) = \pm \frac{1}{4}
$$
we readily arrive at the result. Limits above can be reduced to one of the classic limits involving the exponential function, namely:
$$
    \lim_{t \to 0} \frac{\mathrm{e}^t-1}{t} = 1
$$
Indeed:
$$
   \lim_{y \to 0} \frac{1}{y} \ln(1\pm y) \stackrel{y = \pm \left(\exp(t)-1\right)}{=} \pm \left( \lim_{t \to 0} \frac{t}{\mathrm{e}^t-1}\right) = \pm 1
$$
Using using the substitution $t = \ln \left(\frac{1+\sqrt{1+y^2}}{2}\right)$ we have
$$
 \lim_{y \to 0} \frac{1}{y^2} \ln \left(\frac{1+\sqrt{1+y^2}}{2}\right) = \lim_{t\to 0} \frac{t}{4 \left(\mathrm{e}^{2t} - \mathrm{e}^t\right)} = \frac{1}{4} \cdot \lim_{t \to 0} \mathrm{e}^{-t} \cdot \lim_{t\to 0} \frac{t}{\left(\mathrm{e}^{t} - 1\right)} = \frac{1}{4}
$$
and similarly for the case with $\sqrt{1-y^2}$.
A: It is easy to observe that as $x \to \infty$ the argument of both the logarithms tend to $1$ and this is exactly what we need when we are dealing with limit of logarithmic terms. Just add $1$ and subtract $1$ in the argument of log and proceed as follows:
\begin{align}
L &= \lim_{x \to \infty}\log\frac{x + \sqrt{x^{2} + 1}}{x + \sqrt{x^{2} - 1}}\left(\log\frac{x + 1}{x - 1}\right)^{-2}\notag\\
&= \lim_{t \to 0^{+}}\log\frac{1 + \sqrt{1 + t^{2}}}{1 + \sqrt{1 - t^{2}}}\left(\log \frac{1 + t}{1 - t}\right)^{-2}\text{ (putting }x = 1/t)\notag\\
&= \lim_{t \to 0^{+}}\log\left(1 + \frac{1 + \sqrt{1 + t^{2}}}{1 + \sqrt{1 - t^{2}}} - 1\right)\left\{\log\left(1 + \frac{1 + t}{1 - t} - 1\right)\right\}^{-2}\notag\\
&= \lim_{t \to 0^{+}}\log\left(1 + \frac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}\right)\left\{\log\left(1 + \frac{2t}{1 - t}\right)\right\}^{-2}\notag\\
&= \lim_{t \to 0^{+}}\dfrac{\log\left(1 + \dfrac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}\right)}{\dfrac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}}\cdot\dfrac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}\left\{\log\left(1 + \frac{2t}{1 - t}\right)\right\}^{-2}\notag\\
&= \lim_{t \to 0^{+}}\dfrac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}\left\{\dfrac{\log\left(1 + \dfrac{2t}{1 - t}\right)}{\dfrac{2t}{1 - t}}\cdot\dfrac{2t}{1 - t}\right\}^{-2}\notag\\
&= \lim_{t \to 0^{+}}\dfrac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{1 + \sqrt{1 - t^{2}}}\cdot\frac{(1 - t)^{2}}{4t^{2}}\notag\\
&= \frac{1}{8}\lim_{t \to 0^{+}}\frac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{t^{2}}\notag\\
&= \frac{1}{8}\lim_{t \to 0^{+}}\frac{2t^{2}}{t^{2}\left\{\sqrt{1 + t^{2}} + \sqrt{1 - t^{2}}\right\}}\notag\\
&= \frac{1}{8}\notag
\end{align}
The technique used above is based on the standard limits and is well explained in the comments to this answer.
