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?

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