$\displaystyle\lim_{x \to 0}\left[\dfrac{\log|2 + x| - \log|2 - x|}{\tan x}\right] = $ .......

As $x \to 0$, $2+x > 0$ and $2-x > 0$, so the absolute value signs can be removed:

$\lim_{x \to 0} \dfrac{\log(2 + x) - \log(2 - x)}{\tan x}$

This limit is of the form $\dfrac{0}{0}$. Applying L'Hospital's rule by differentiating the numerator and the denominator with respect to $x$:

$\lim_{x \to 0} \dfrac{\dfrac{1}{2 + x} - \dfrac{1}{2 - x}(-1)}{\sec^2 x}$

$= \lim_{x \to 0} \dfrac{\dfrac{1}{2 + x} + \dfrac{1}{2 - x}}{\sec^2 x}$

Substituting $x = 0$:

$= \dfrac{\dfrac{1}{2} + \dfrac{1}{2}}{1} = 1$

Answer: $1$