MathematicsMedium179×since 2002Q4226limx→0 (27+x)13−39−(27+x)23\mathop {\lim }\limits_{x \to 0} \,\,{{{{\left( {27 + x} \right)}^{{1 \over 3}}} - 3} \over {9 - {{\left( {27 + x} \right)}^{{2 \over 3}}}}}x→0lim9−(27+x)32(27+x)31−3 equals.A13{1 \over 3}31B−-− 13{1 \over 3}31C−-− 16{1 \over 6}61D16{1 \over 6}61Check answerSkip