$\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\frac{8-{{x}^{3}}}{\left| 3{{x}^{2}}-13x+14 \right|}\underline{\underline{\frac{0}{0}}}\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\frac{(2-x)(4+2x+{{x}^{2}})}{\left| (x-2)(3x-7) \right|}=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\frac{(2-x)(4+2x+{{x}^{2}})}{-(x-2)(3x-7)}=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\frac{4+2x+{{x}^{2}}}{3x-7}=-12$