For each $t \in R$ let $[t]$ be the greatest integer less than or equal to t. Then, $\mathop {\lim }\limits_{x \to 1 + } \frac{{(1 - \left| x \right| + \sin \left| {1 - x} \right|\sin \left( {\frac{\pi }{2}[1 - x]} \right)}}{{\left| {1 - x} \right|[1 - x]}}$

JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Analyzing +4 | -1

For each

t \in R

let

[t]

be the greatest integer less than or equal to t. Then,

\lim\limits_{x \to 1 + } \frac{{(1 - \left| x \right| + \sin \left| {1 - x} \right|\sin \left( {\frac{\pi }{2}[1 - x]} \right)}}{{\left| {1 - x} \right|[1 - x]}}

View answer and explanation

Correct option: A

equals 0

Option A (equals 0) is correct because after substituting x = 1 + h and simplifying, the limit evaluates to 0.