applying limit on Floor Function

Question

Applying limit on a normal function is fine but I have never applied limit on floor, how do we find the following limit $$\lim_{x\to\infty}\frac{\lfloor{\frac{0.01}{x}}\rfloor}{\log x}$$

Answer 1

You can rewrite the limit by changing the variable as $y=\frac{0.01}{x}$ and so $x = \frac{0.01}{y}$. As $x$ goes to infinity, $y$ goes to $0^+$.

$$\lim_{x\to \infty} \frac{\lfloor\frac{0.01}{x}\rfloor}{log x}=\lim_{y\to 0^+} \frac{\lfloor y \rfloor}{log \frac{0.01}{y}}$$

Now you need to calculate the limit at zero from the right.

$$\lim_{y\to 0^+} \frac{\lfloor y \rfloor}{log \frac{0.01}{y}}=\lim_{y\to 0^+} \frac{0 (absolute \, zero)}{a \, positive \, value} = 0$$

Therefore, it can be said that $\lim_{x\to \infty} \frac{\lfloor\frac{0.01}{x}\rfloor}{log x} = 0$.

Answer 2

Sorry but I have different opinion with Kian. I think the limit exists and it is $0$.
First, I talk about how I solve this problem:
First make restriction that $x>0.01$ so that the numerator is $0$, a constant which has nothing to do with $x$. So the function become a constant function. So the limit is $0$.

Second I talk about the problem in Kian's answer: He thinks that $x$ goes to infinity equals to $y$ goes to $0$ in two sides, actually I think it is not right. $y=0.01/x$ , $x$ increase from $0+$ to infinity, so $y$ decrease from infinity to $0+$. I think, we only need to consider the one side.