If l1=limx→−2 (x+|x|),l2=limx→−2 (2x+|x|) and l3=limx→π/2 cos⁡xx−π/2, then 

If l1=limx2(x+|x|),l2=limx2(2x+|x|) and l3=limxπ/2cosxxπ/2, then 

  1. A

    l1<l2<l3

  2. B

    l2<l3<l1

  3. C

    l3>l2>l1

  4. D

    l1<l3<l2

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    Solution:

    We have, 

    l1=limx2(x+|x|)=2+2=0l2=limx2(2x+|x|)=4+2=2

    and, l3=limxπ/2cosxxπ/2=limxπ/2sin(π/2x)(π/2x)=1

              I2<l3<I1.

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