A limit with floor function

Let $\left \lfloor \cdot \right \rfloor$ denote the floor function. Prove that:

$$\lim_{x\rightarrow 0} \; x \left \lfloor \frac{1}{x} \right \rfloor=1$$

Solution

It is well known that forall $t \in \mathbb{R}$ it holds that $0\leq t - \left \lfloor t \right \rfloor <1$. Thus for $x \neq 0$ we get that:


\begin{align*}
0\leq \frac{1}{x} - \left \lfloor \frac{1}{x} \right \rfloor <1 &\Rightarrow - \frac{1}{x} \leq - \left \lfloor \frac{1}{x} \right \rfloor < 1- \frac{1}{x}\\
 &\Rightarrow -1 \leq -x \left \lfloor \frac{1}{x} \right \rfloor<x - 1 \\
 &\Rightarrow 1-x < x \left \lfloor \frac{1}{x} \right \rfloor \leq 1
\end{align*}

Now for $x>0$ taking limit as $x \rightarrow 0+$ at the equation above we get that the limit is $1$, while for $x<0$ we have that $1\leq x \left  \lfloor \frac{1}{x} \right \rfloor <1-x$ and by taking limit as $x \rightarrow 0-$ we get that the limit is also $1$. Hence:

$$\lim_{x\rightarrow 0} \; x \left \lfloor \frac{1}{x} \right \rfloor=1$$

The exercise can also be found here