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For each `tinR," let "[t]` be the greatest integer less than or equal to t. Then `lim_(xto0^(+)) x([(1)/(x)]+[(2)/(x)]+...+[(15)/(x)])`

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The correct Answer is:
D
`underset(xto0^(+))lim([(1)/(x)]+[(2)/(x)]+...+[(15)/(x)])`
`=underset(xto0^(+))lim(((1)/(x)+(2)/(x)+(3)/(x)+...+(15)/(x))-({(1)/(x)}+{(2)/(x)}+...+{(15)/(x)}))`
(Where `{k}` represents fractional part of k)
`=underset(xto0^(+))lim(x(120)/(x)-x({(1)/(x)}+{(2)/(x)}+...+{(15)/(x)}))`
`=120-0" "(because0lt{k}lt1)`
`=120`
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