Answer
$1$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to 0^{+}} \dfrac{ \ln (\sin x)}{\ln (\tan x)}$.
But $\lim\limits_{x \to 0^{+}} \dfrac{ \ln (\sin x)}{\ln (\tan x)}=\dfrac{\infty}{\infty}$
We can see that the numerator and denominator have a limit of $\infty$, so the limit shows the indeterminate form of type $\dfrac{\infty}{\infty}$. So, we will apply L'Hopital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$
where, $a$ can be any real number, infinity or negative infinity.
$\lim\limits_{x \to 0^{+}} \dfrac{ \dfrac{d}{dx} \ln (\sin x)}{\dfrac{d}{dx} \ln (\tan x)}=\lim\limits_{x \to 0^{+}} \dfrac{(1/\sin x) \cos x}{(1/\tan x) \sec^2 x}\\ =\lim\limits_{x \to 0^{+}} \dfrac{\cot x}{\cot x \sec^2 x}=1$
