Answer
$0$
Work Step by Step
Here, we have $\lim\limits_{x \to 0^{+}} f(x)=\lim\limits_{x \to 0^{+}} \dfrac{\ln x}{\csc x}$
But $\lim\limits_{x \to 0^{+}} \dfrac{\ln x}{\csc x}=\dfrac{\infty}{-\infty}$
This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$\lim\limits_{x \to 0^{+}} \dfrac{1/x}{-\csc x \cot x}=\lim\limits_{x \to 0^{+}} \dfrac{-\sin x \tan x}{x}$
or, $ \dfrac{-\sin 0 \tan 0}{0}=\dfrac{0}{0}$
This shows an indeterminate form of limit, thus we will again apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$\lim\limits_{x \to 0^{+}} \dfrac{(-\cos x)(\tan x)+(-\sin x)(\sec^2 x)}{1}=\dfrac{(-1)(0)+(0)(1)}{1}$
and $\dfrac{0}{1} =0$
