Limit of logarithm of square roots

$\begin{aligned} L & = \lim_{x \to \infty} \log_x \left(\sqrt{x^2+1} - \sqrt{x^2-1} \right) \\ & = \lim_{x \to \infty} \frac {\ln \left(\sqrt{x^2+1} - \sqrt{x^2-1} \right)}{\ln x} & \small {\color{#3D99F6}\text{A }\infty/\infty \text{ case, L'Hôpital's rule applies.}} \\ & = \lim_{x \to \infty} \frac {-\frac x{\sqrt{(x^2+1)(x^2-1)}}}{\frac 1x} & \small {\color{#3D99F6}\text{Differentiate up and down w.r.t. }x} \\ & = \lim_{x \to \infty} -\frac {x^2}{\sqrt{x^4-1}} & \small {\color{#3D99F6}\text{Divide up and down by }x^2} \\ & = \lim_{x \to \infty} -\frac 1{\sqrt{1-\frac 1{x^4}}} \\ & = \boxed{-1} \end{aligned}$

$\begin{aligned} & \lim_{x\to\infty} \log_x \left(\sqrt{x^2+1} - \sqrt{x^2-1}\right)\\ = & \lim_{x\to\infty} \log_x \left(\dfrac{2}{\sqrt{x^2+1}+\sqrt{x^2-1}} \right)\\ = & \lim_{x\to\infty} \left[\log_x(2) - \log_x(x) - \log_x\left(\sqrt{1+\dfrac{1}{x^2}} + \sqrt{1-\dfrac{1}{x^2}} \right) \right]\\ = & -1 \end{aligned}$