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 ${l_3} = \mathop {\lim }\limits_{x \to \pi /2} \frac{{\cos x}}{{x - \pi /2}}$, then

JEE MATHEMATICS MCQ: LIMITS

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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}}{{x - \pi /2}}

, then

View answer and explanation

Correct option: B

{l_2} < {l_3} < {l_1}

Option B (

{l_2} < {l_3} < {l_1}

) is correct because it accurately compares the limits calculated by considering the absolute value and the limit of the trigonometric function.