Answer
$0$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to 0^{+}} \tan x \ln x=\lim\limits_{x \to 0^{+}} \dfrac{\ln x}{1/\tan x}$.
But $\lim\limits_{x \to 0^{+}} \dfrac{\ln x}{1/\tan x}=\dfrac{\infty}{\infty}$
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.
$\lim\limits_{x \to 0^{+}} \dfrac{\ln x}{1/\tan x}=\lim\limits_{x \to 0^{+}} \dfrac{1/x}{-\csc^2 x} \\ =- \lim\limits_{x \to 0^{+}} \dfrac{\sin^2 x}{x^2} \times x \\=-(1)^2 \times 0 \\=0$