Can we say that if $x\to 0^+$ then $\left \lfloor{\frac{1}{x}}\right \rfloor \sim \frac{1}{x}$?

Question

Can we say that if $x\to 0^+$ then $\left \lfloor{\frac{1}{x}}\right \rfloor \sim \frac{1}{x}$?

Since $x\to 0^+\implies \frac{1}{x}\to +\infty$ therefore $\left \lfloor{\frac{1}{x}}\right \rfloor \to +\infty$.

So can we conclude that $\left \lfloor{\frac{1}{x}}\right \rfloor \sim \frac{1}{x}$ as $x\to 0^+$?

Thanks for your help.

Answer 1

The fact that both go to infinity is not enough to conclude that they are asymptotically equivalent. For example, $\frac{1}{x}$ and $\frac{1}{x^2}$ a lso both go to $\infty$ as $x\rightarrow0^+$, but their ratio goes to $0$. However, $$\lim_{x\rightarrow0^+}\frac{\frac{1}{x}}{\left\lfloor\frac{1}{x}\right\rfloor}=\lim_{x\rightarrow0^+}\frac{\left\lfloor\frac{1}{x}\right\rfloor+\left\{\frac{1}{x}\right\}}{\left\lfloor\frac{1}{x}\right\rfloor}=\lim_{x\rightarrow0^+}\left(1+\frac{\left\{\frac{1}{x}\right\}}{\left\lfloor\frac{1}{x}\right\rfloor}\right)=1.$$ The last equality follows from the squeeze theorem, since $$0\le\frac{\left\{\frac{1}{x}\right\}}{\left\lfloor\frac{1}{x}\right\rfloor}\le\frac{1}{\left\lfloor\frac{1}{x}\right\rfloor}\stackrel{x\rightarrow0^+}{\rightarrow}0.$$ Therefore, $\frac{1}{x}\sim\left\lfloor\frac{1}{x}\right\rfloor$ as $x\rightarrow0^+$.

Answer 2

Your argument is not sufficient for concluding that. The following is the complete argument: $$\lim_{x\to0}x\left(\frac{1}{x}-1\right)<\lim_{x\to0}x\left\lfloor\frac{1}{x}\right\rfloor\le\lim_{x\to0}x\times\frac{1}{x}$$ Now, the result follows from sandwich theorem.