JEE Main 201910 Jan, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002For each t ∈\in∈ R , let [t] be the greatest integer less than or equal to t Then limx→1+(1−∣x∣+sin∣1−x∣)sin(π2[1−x])∣1−x∣.[1−x]\mathop {\lim }\limits_{x \to 1^ + } {{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {{\pi \over 2}\left[ {1 - x} \right]} \right)} \over {\left| {1 - x} \right|.\left[ {1 - x} \right]}}x→1+lim∣1−x∣.[1−x](1−∣x∣+sin∣1−x∣)sin(2π[1−x])AOption A. equals −-− 1BOption B. equals 1COption C. equals 0DOption D. does not existCheckSkip · show solution
