limy→0 1+1+y4−2y4

limy01+1+y42y4

  1. A

    exists and equals 142

  2. B

    does not exist

  3. C

    exists and equals 122

  4. D

    exists and equals 122(2+1)

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    Solution:

       limy01+1+y42y4

    =limy01+1+y42y41+1+y4+2=limy01+y411+y4+1y41+1+y4+21+y4+1

    =limy01+y41y41+1+y4+21+y4+1=limy011+1+y4+21+y4+1=142

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