Bounds of $f(x)= \frac{x-\lfloor x\rfloor}{\sqrt x}$

Question

Let $f$ be defined by:

$f(x)= \frac{x-\lfloor x\rfloor}{\sqrt x}$

1. Prove that $f$ is bounded in $\mathbb{R+}$

I have no idea where to start, I can see by graphing it here (desmos) that it is bounded by $0$ and $1$, but here we are not presented with a graph.

Answer 1

Hints:

• for all $\,x \in \mathbb{R}^+\,$ you have $0 \le x - \lfloor x \rfloor \lt 1\,$, so $\,0 \le f(x) \lt \frac{1}{\sqrt{x}}$

• for $x \in (0,1)\,$ you have $\,\lfloor x \rfloor = 0\,$, so $\;f(x) = \sqrt{x} \lt 1$

Answer 2

To prove that a function is bounded you need to use inequalities.

Note that

$$x\in(0,1) \quad f(x)= \frac{x-\lfloor x\rfloor}{\sqrt x}\leq\frac{1}{\sqrt x}$$

$$\forall x\geq 0 \quad f(x)= \frac{x}{\sqrt x}={\sqrt x}\leq1$$

$\implies$ $\forall x \in \mathbb{R^+}$ f is bounded.