I've been trying to solve the following limit:
$$ \lim_{{x \to 0^+}} \sqrt{\frac{e^{\Gamma(x)} - e^{-\Gamma(x)}}{\ln(\Gamma(x))}}$$
where $\Gamma(x)$ is the Gamma function and:
$$ \lim_{{x \to 0^+}} \Gamma(x) = \infty$$
Upon using L'Hopital, I arrived at the $\infty/\infty$ indeterminate form, and further calculations involved derivatives in droves that only made the case more convoluted. I've also tried putting the exponentials in evidence:
$$\lim_{{x \to 0^+}} \sqrt{\frac{e^{\Gamma(x)} (1- e^{-2\Gamma(x)})}{\ln(\Gamma(x))}} = \lim_{{x \to 0^+}} \sqrt{\frac{e^{\Gamma(x)} }{\ln(\Gamma(x))}} \lim_{{x \to 0^+}} \sqrt{(1- e^{-2\Gamma(x)})} = \lim_{{x \to 0^+}} \sqrt{\frac{e^{\Gamma(x)} }{\ln(\Gamma(x))}} \hspace{0,5cm}, $$
albeit, this is only possible if the product of the limits doesn't lead to an indeterminate form, so the attempt above is wrong.
I would argue that the exponential grows substantially faster than the logarithm in the denominator, but still, I'm faced with a contrivance in that we still get a $\infty / \infty$ there. How do I proceed?