Decrease Impedance at Higher Frequencies
"since they are in parallel Xtot should decrease"


That's not automatically true for complex impedances. For the inductor \$Z_L = j\omega L\$, for the capacitor \$Z_C = \dfrac{1}{j \omega C}\$. Both in parallel gives  

\$ Z = \dfrac{j\omega L \cdot \dfrac{1}{j \omega C}}{j\omega L + \dfrac{1}{j \omega C}} = \dfrac{j\omega L \cdot \dfrac{1}{j \omega C}}{\dfrac{j^2 \omega^2 LC + 1}{j \omega C}} = \dfrac{j\omega L}{1 - \omega^2 LC} \$

The minus sign in the denominator is interesting. Since \$\omega^2 LC\$ is positive, we can find a value for it where the denominator becomes zero.

\$ \omega^2 LC = 1\$

or 

\$ \omega = \sqrt{\dfrac{1}{LC}} \$

If the denominator is zero the impedance is infinite. That goes against intuition which says that paralleling two components will give a lower impedance than the lowest of the two.   

  

The image gives the explanation. Currents through \$C\$ and \$L\$ are at 90° with voltage, but in opposite directions. If their magnitudes are different their sum will be either a capacitive or inductive current. But when they're equal the sum is zero. No current. A zero current for a non-zero voltage means infinite impedance.  

The frequency for which this is true is the resonance frequency, and it's used to make oscillators and high-Q filters.