JEE Main 2018Session, Shift 1MathematicsHardLimit, Continuity and Differentiability420× since 2002For each t ∈R\in R∈R, let [t] be the greatest integer less than or equal to t. Then limx→0+x([1x]+[2x]+.....+[15x])\mathop {\lim }\limits_{x \to {0^ + }} x\left( {\left[ {{1 \over x}} \right] + \left[ {{2 \over x}} \right] + ..... + \left[ {{{15} \over x}} \right]} \right)x→0+limx([x1]+[x2]+.....+[x15])AOption A. does not exist in RBOption B. is equal to 0COption C. is equal to 15DOption D. is equal to 120CheckSkip · show solution
