JEE Main 201816 Apr, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002limx→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.AOption A. 13{1 \over 3}31BOption B. −-− 13{1 \over 3}31COption C. −-− 16{1 \over 6}61DOption D. 16{1 \over 6}61CheckSkip · show solution
