Answer
$-\infty$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to 0^{+}} \dfrac{\cot x}{\ln (x)}$.
But $\lim\limits_{x \to 0^{+}} \dfrac{\cot x}{\ln (x)}=\dfrac{\infty}{\infty}$
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: $\lim\limits_{x \to a} \dfrac{A(x) }{B(x)}=\lim\limits_{x \to a}\dfrac{A'(x)}{B'(x)}$
where $a$ can be any real number, infinity or negative infinity.
$\lim\limits_{x \to 0^{+}} \dfrac{\cot x}{\ln (x)}=\lim\limits_{x \to 0^{+}} \dfrac{-1/\sin^2 x}{1/x} \\=- \lim\limits_{x \to 0^{+}} \dfrac{1}{\dfrac{\sin^2 x}{x}}\\=- \lim\limits_{x \to 0^{+}} \dfrac{1}{\dfrac{\sin x}{x}} \times \lim\limits_{x \to 0^{+}} \dfrac{1}{\sin x} \\=(-1) \times (+\infty)\\=-\infty$