How to compute the following limit? $\lim\limits_{x\to\ 0^+} {\frac{x-\lfloor x \rfloor}{x+\lfloor x \rfloor}}$

Question

$$\lim\limits_{x\to\ 0^+} {\frac{x-\lfloor x \rfloor}{x+\lfloor x \rfloor}}$$ Here, $\lfloor x \rfloor$ represents the floor of $x$.

I tried using a graphing calculator (desmos) to plot the function $f(x) = \frac{x-\lfloor x \rfloor}{x+\lfloor x \rfloor}$

from the graph (see image here), it is clear that this limit is equal to 1. But is there any analytical way to compute this limit?

Answer 1

For $0<x<1$, you have that $\lfloor x\rfloor =0$. Hence $$ \lim\limits_{x\to\ 0^+} {\frac{x-\lfloor x \rfloor}{x+\lfloor x \rfloor}} =\lim\limits_{x\to\ 0^+} \frac{x}{x}=1. $$

Answer 2

From the graph, it's not only clear that the limit is $1$, it can also be seen that on the interval $[0,1)$, the vaue of $f(x)$ is equal to $1$. This can easily be shown analytically, since for $x\in[0,1)$, you have $\lfloor x\rfloor = 0$.