$$\lim_{x \to \infty}\ln{\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}}}\cdot \left(\ln{\frac{x+1}{x-1}}\right)^{-2}=\frac{1}{8}$$
Any suggestion to find this limit without series expansion and l'Hôpital's rule? Thanks and regards.
Note: WolframAlpha confirms that the result is $\frac{1}{8}$.
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 $\sqrt{1-y^2}$.