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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?

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  • $\begingroup$ Thank your for the observation. $\endgroup$ – user926356 Sep 3 '20 at 6:12
  • $\begingroup$ Yes. That is correct. $\endgroup$ – steven gregory Sep 13 '20 at 5:06
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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 $\boxed{1}$

And yes, your arguement is also correct

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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 $0\le\{x\}<1$

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How about

$ \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$

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