We have
$\frac{ \lfloor x\rfloor}{x} = \frac{ x- y}{x}$ where y is fractional part of x and hence $0 \le y \lt 1$
Hence
$ \frac{ \lfloor x\rfloor}{x} = 1- \frac{y}{x} \ge 1 - \frac{1}{x}$
as x goes to infinitte RHS goes to 1 so the required value is between 1 and 1 so it is 1
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