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How can I prove that $$\lim_{x\to +\infty}\frac{\left \lfloor{x}\right \rfloor}{x}=1$$

L'Hôpital's rule seems to fail here, since the floor function is not differentiable for integers. What other ways are there to prove this?

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    $\begingroup$ Try squeezing it. $\endgroup$ – kingW3 Jun 20 '17 at 17:17
  • $\begingroup$ yes you can squeeze it between for example x+1 and x-1 in numerator $\endgroup$ – mathreadler Jun 20 '17 at 17:20
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Note that $$ x=\lfloor{x}\rfloor+\zeta $$ where $0\leq \zeta<1.$ Then $$ \frac{x}{\lfloor{x}\rfloor}=1+\frac{\zeta}{\lfloor{x}\rfloor}\to1 $$ as $x\to \infty$ since $\zeta/\lfloor{x}\rfloor\to 0$ as $x\to\infty$.

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  • $\begingroup$ You swapped nominator and denominator with respect to the question, otherwise fine. $\endgroup$ – M. Winter Jun 20 '17 at 17:30
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Big hint: $x-1 \le \lfloor x \rfloor \le x$ for all $x \in \Bbb R$.

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