limx→0 2|x|e|x|−|x|−|x|loge⁡2−1xtan⁡x is equal to

 limx02|x|e|x||x||x|loge21xtanx is equal to

  1. A

    12(ln2)2+12(ln2)+1

  2. B

    (ln2)2+12(ln2)+1

  3. C

    (ln2)2+(ln2)+12

  4. D

    12(ln2)2+(ln2)+12

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    Solution:

         limx02|x|e|x||x||x|loge21xtanx

    =limx0(2e)|x||x|loge(2e)1x2tanxx=limx01+|x|loge(2e)+|x|22!loge(2e)2+|x|loge(2e)1x2tanxx

    =limx0|x|22!loge(2e)2+..x2tanxx=12!loge(2e)2=12loge2+12=12(ln2+1)2

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