Can we say that if $x\to 0^+$ then $\left \lfloor{\frac{1}{x}}\right \rfloor \sim \frac{1}{x}$? 
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.
 A: 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 $x\rightarrow0^+$.
A: 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.
