Answer
$0$
Work Step by Step
L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$
Here, $\lim\limits_{x \to 0^{+}} f(x)=\lim\limits_{x \to 0^{+}} \dfrac{\ln x}{\csc x}$
Now, $\lim\limits_{x \to 0^{+}} \dfrac{\ln x}{\csc x}=\dfrac{\infty}{-\infty}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule. rm of limit so we need to use L-Hospital's rule.
$\lim\limits_{x \to 0^{+}} \dfrac{1/x}{-\csc x \cot x}=\lim\limits_{x \to 0^{+}} \dfrac{-\sin x \tan x}{x} =\dfrac{0}{0}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$\lim\limits_{x \to 0^{+}} \dfrac{(-\cos x)(\tan x)+(-\sin x)(\sec^2 x)}{1}=\dfrac{(-1)(0)+(0)(1)}{1} =0$