# Find a limit involving floor function

I had to find the following limit: $$\displaystyle{\lim_{x \to \infty}}\frac{x}{\lfloor x \rfloor}$$ Where $$x\in\mathbb{R}$$ and $$f(x)= {\lfloor x \rfloor}$$ denotes the floor function.

This is what I did:

I wrote $$x-1\leq \lfloor x \rfloor\leq x$$. Then $$\frac{1}{x}\leq\frac{1}{\lfloor x \rfloor}\leq\frac{1}{x-1}$$. Multiplying by $$x$$ we get $$\frac{x}{x}\leq\frac{x}{\lfloor x \rfloor}\leq\frac{x}{x-1}$$. Since $$\displaystyle{\lim_{x \to \infty}}\frac{x}{x}=\displaystyle{\lim_{x \to \infty}}\frac{x}{x-1}=1$$, by the squeeze theorem we can say that $$\displaystyle{\lim_{x \to \infty}}\frac{x}{\lfloor x \rfloor}=1$$. Is this correct? If not, what should I fix?

• Thank your for the observation. Sep 3, 2020 at 6:12
• Yes. That is correct. Sep 13, 2020 at 5:06

I'd rather convert it to $$\lim_{x\to \infty}\dfrac{x}{x-\lbrace x \rbrace}=\lim_{x\to \infty}\dfrac{1}{1-\boxed{\dfrac{\lbrace x \rbrace}{x}}}$$ $$\dfrac{\lbrace x \rbrace}{x}\to0$$since $$\lbrace x \rbrace$$ is just a number $$\in[0,1)$$
$$\therefore$$ the limit is $$\boxed{1}$$
It is correct. Slightly easier $$\lfloor x\rfloor = x-\{x\}$$ So $$x/\lfloor x \rfloor= \frac{1}{1-\{x\}/x}$$ so the limit is $$1$$ since $$0\le\{x\}<1$$
$$\displaystyle{\lim_{x \to \infty}}\frac{x}{\lfloor x \rfloor} = {\lim_{x \to \infty}}\frac{\lfloor x \rfloor + \{x\}}{\lfloor x \rfloor} = 1+{\lim_{x \to \infty}}\frac{\{x\}}{\lfloor x \rfloor} = 1$$