Step 1: Rewrite the given expression as an exponential.
Using $x=e^{\ln x}$, we have $x^{\log_e x}=e^{(\log_e x)^2}$.
Step 2: Read what the limit forces.
We are told $\displaystyle\lim_{x\to\infty}e^{(\log_e x)^2}=0$. An exponential tends to $0$ only when its exponent tends to $-\infty$.
Step 3: Interpret the exponent condition.
A square $(\log_e x)^2$ cannot be negative for ordinary real values, so the only way this special condition holds is with $\log_e x<0$, that is $0<x<1$.
Step 4: Express the target with change of base.
$\log_x 12=\dfrac{\log_e 12}{\log_e x}$.
Step 5: Decide the sign of each part.
Since $12>1$, the numerator $\log_e 12>0$. From Step 3, the denominator $\log_e x<0$.
Step 6: Combine the signs.
A positive number divided by a negative number is negative, so $\log_x 12<0$, i.e. Negative.
\[ \boxed{\text{Negative}} \]