If ${l_1} = \mathop {\lim }\limits_{x \to {2^ + }} \left( {x + \left[ x \right]} \right),{l_2} = \mathop {\lim }\limits_{x \to {2^ - }} \left( {2x - \left[ x \right]} \right)$ and ${l_3} = \mathop {\lim }\limits_{x \to \pi /2} \frac{{\cos x}}{{\left( {x - \pi /2} \right)}}$, then

JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

{l_1} = \lim\limits_{x \to {2^ + }} \left( {x + \left[ x \right]} \right),{l_2} = \lim\limits_{x \to {2^ - }} \left( {2x - \left[ x \right]} \right)

and

{l_3} = \lim\limits_{x \to \pi /2} \frac{{\cos x}}{{\left( {x - \pi /2} \right)}}

, then

View answer and explanation

Correct option: C

{l_3} < {l_2} < {l_1}

Option C (

{l_3} < {l_2} < {l_1}

) is correct because evaluating each limit separately gives l1 = 4, l2 = 3, and l3 = -1, which satisfy the inequality.