Limit of $\frac{\sqrt{|x |-\lfloor x\rfloor }}{x^2}$ using only definition

Question

How can I show that

$$\lim _{|x|\to \infty \:}\frac{\sqrt{\left|x\right|-\lfloor \:x\rfloor }}{x^2}=0$$

by using only definition of limit

$$ \lim _{x\to +\infty}f(x)=L \iff \forall \varepsilon >0\quad \exists A >0\quad \forall x\in U\quad x> A \Rightarrow |f(x)-L|<\varepsilon $$

• As first step I tired to show that

$$\left|\frac{\sqrt{\left|x\right|-\lfloor \:x\rfloor }}{x^2} -0 \right| \leq \frac{1}{x^2}$$ by with no luck indeed,

$$ x-1 <\lfloor x \rfloor \leq x < \lfloor x \rfloor+1 $$ $$x-1 <\lfloor x \rfloor \leq x \implies -x \leq -\lfloor x \rfloor <|x|+x-1 \implies |x|-x \leq |x|-\lfloor x \rfloor <x-1 $$

• Second step we have for all $x >0$ $$\left|\frac{\sqrt{\left|x\right|-\lfloor \:x\rfloor }}{x^2} -0 \right| \leq \frac{1}{x^2}$$ Let $\varepsilon >0$

Answer 1

We have that $|x|^2=x^2$ and $\lfloor \:x\rfloor \leq x \leq |x|$

Thus $$\frac{\sqrt{\left|x\right|-\lfloor \:x\rfloor }}{x^2}\leq \frac{\sqrt{|x|+|x|}}{x^2}\leq \sqrt{2} \frac{\sqrt{|x|}}{|x|^2} \leq \frac{2}{|x|^{\frac{3}{2}}}$$

Put $a=\frac{3}{2}$

We have that $\lim_{x \to +\infty}\frac{2}{|x|^a}=0$

Thus from the definition we must find $S>0$ such that $\frac{1}{|x|^a}<\epsilon ,\forall x >S$

Now $$\frac{1}{|x|^a}<\epsilon \Rightarrow |x|^a> \frac{1}{\epsilon} \Rightarrow |x|>\sqrt[a]{(\frac{1}{\epsilon})}$$

Take $S=\sqrt[a]{(\frac{1}{\epsilon})}$ and we have that if $|x|>S$ then $\frac{1}{|x|^a}<\epsilon \Rightarrow \frac{\sqrt{\left|x\right|-\lfloor \:x\rfloor }}{x^2}< \epsilon$

Answer 2

Hint

$$\left|\frac{\sqrt{\left|x\right|-\lfloor \:x\rfloor }}{x^2} -0 \right| \leq \frac{1}{x^2}$$

Now to make $\frac{1}{x^2} <\epsilon$ you need $|x| >...$.

Answer 3

The notation $\lim_{|x|\to\infty}$ is not standard but I guess the intended meaning is via this definition.

Let $f(x)$ be defined for all $x$ with $|x|> a$. The notation $\lim_{|x|\to\infty}f(x)=L$ means that corresponding to every $\epsilon > 0$ there is an $N> 0$ such that $|f(x) -L | < \epsilon$ whenever $|x|>N$.

It is easy to note that $x - 1 < \lfloor x\rfloor\leq x$ and $|x|=\max(x, -x)$ so that $0\leq |x| - \lfloor x\rfloor \leq 2|x| + 1$ and now it is easily seen via Sqeeze that the desired limit is $0$.