One of the possible functions \(f(x)\) which satisfies \(\int\limits_{-2}^{2} f(x)\,dx = 0\) is

For an odd function \(f(x)\), the definite integral over a symmetric interval is zero, i.e., \(\int_{-a}^{a} f(x)\,dx = 0\).

Let us check the function in the first option:

\(f(x) = \log\left(\dfrac{2 - x}{2 + x}\right)\)

Substituting \(-x\) for \(x\), we get:

\(f(-x) = \log\left(\dfrac{2 - (-x)}{2 + (-x)}\right) = \log\left(\dfrac{2 + x}{2 - x}\right)\)

Using the property of logarithms \(\log(a^{-1}) = -\log(a)\):

\(f(-x) = \log\left(\left(\dfrac{2 - x}{2 + x}\right)^{-1}\right) = -\log\left(\dfrac{2 - x}{2 + x}\right) = -f(x)\)

Since \(f(-x) = -f(x)\), the function \(f(x) = \log\left(\dfrac{2 - x}{2 + x}\right)\) is an odd function.

Therefore, \(\int_{-2}^{2} \log\left(\dfrac{2 - x}{2 + x}\right)\,dx = 0\).

Answer: \(\log\left(\dfrac{2 - x}{2 + x}\right)\)