For each t∈R, let [t] be the greatest integer less than or equal to t. Then,limx→0+ x1x+2x+……+15x

For each tR, let [t] be the greatest integer less than or equal to t. Then,

limx0+x1x+2x++15x

  1. A

    is equal to 15

  2. B

    is equal to 120

  3. C

    does not exist in R

  4. D

    is equal to 0.

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    Solution:

    We know that [x]=x{x}, where {x} denotes the fractional part of x.

    limx0+x1x+2x++15x1x+2x++15x=limx0+(1+2++15)limx0+x1x+2x++15x

    =15(15+1)2+0×A finite number less than 15=120

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