`lim_(xrarr oo) (logx)/([x])` , where `[.]` denotes the greatest integer function, is

Options

A. 0
B. 1
C. -1
D. non -existant

Answer

A

Explanation

We have
`x-1lt[x]ge x " for all " x in R`
` rArr (1)/(x)le (1)/([x])lt (1)/(x-1) " for all " x in R -{0,1}`
` rArr (logx)/(x)le (logx)/([x]) lt (logx)/(x-1)[ because log x gt 0 " as "x to oo]`
` rArr lim_(xto oo) le lim_(xto oo) (log x)/([x])lt lim_(xle oo) (logx)/(x-1)`
` rArr lim_(xto oo) (log x)/([x]) =0 [because lim_(xto oo) (logx)/(x) =lim_(xto oo) (log x)/(x-1)=0]`