# Limit of quotient involving floor function and identity function

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?

• Try squeezing it. – kingW3 Jun 20 '17 at 17:17
• yes you can squeeze it between for example x+1 and x-1 in numerator – mathreadler Jun 20 '17 at 17:20

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 $x\to\infty$.
Big hint: $x-1 \le \lfloor x \rfloor \le x$ for all $x \in \Bbb R$.