Find the limit of a $f(x)=\frac{\lfloor x^2\rfloor}{x^2}$ at an arbitrary point

Question

The function f is defined $f(x)=\frac{\lfloor x^2\rfloor}{x^2}$ I need to find the limit of the function at an arbitrary point.

For the continuous parts it was fine, and also for right sided limit at positive points of discontinuity (and left sided for negatives, for all of which the lim is 1), and now I'm left with left sided limit of the function at positive points of discontinuity (and vice versa for the negative part).

I know the answer from intuition: $\frac{(x^2-1)}{x^2}$, but I can't find the key to the proof.

My attempt is as follows (letting the point of discontinuity be $x_0>0$)

First, I restrict delta such that $f(x)=\frac{(x_0^2-1)}{x^2}$

Then, if $x_0=1$ then $f(x)-L=0< \varepsilon$ (restricting $x>0$)

Otherwise, I restrict my neighborhood again to $1< x< x_0$, such that

$x<x_0 \rightarrow ... \rightarrow 1/x^2 - 1/x_0^2$

and therefore $f(x)-L=...=(x_0^2-1)(1/x^2 - 1/x_0^2)< \varepsilon$

And now I'm stuck... I can't seem to find the right combo to make the delta-epsilon magic to work.

Thanks for any help!

Answer 1

Note: We know the floor function $\lfloor x\rfloor$ fulfills $$\lfloor x\rfloor \leq x < \lfloor x\rfloor+1$$ with points of discontinuity at the integers. So, its convenient to partition the set $\mathbb{R}$ of real numbers according to the points of discontinuity of the function \begin{align*} &f:\mathbb{R}\backslash\{0\}\rightarrow\mathbb{R}\\ &f(x)=\frac{\lfloor x^2\rfloor}{x^2} \end{align*} which is defined for all real numbers except $0$. We set $$\mathbb{R}\backslash\{0\}=(-\infty,0)\cup[0,\infty)\backslash\{0\}$$ and \begin{align*} [0,\infty)&=\bigcup_{n\geq 0}A_n\qquad \text{ with }A_n=[\sqrt{n},\sqrt{n+1})\\ (-\infty,0)&=\bigcup_{n\geq 1}B_n\qquad \text{ with }B_n=[-\sqrt{n},-\sqrt{n-1}) \end{align*} The set $\{A_n,B_{n+1}|n\geq 0\}$ forms a partition of pairwise disjoint, non-empty, right half-open intervals with union equal $\mathbb{R}$ (and with different lengths). \begin{align*} \mathbb{R}&=\ldots\cup[-2,-\sqrt{3})\cup[-\sqrt{3},-\sqrt{2})\cup[-\sqrt{2},-1)\cup[-1,0)\\ &\qquad\qquad\cup[0,1)\cup[1,\sqrt{2})\cup[\sqrt{2},\sqrt{3})\cup[\sqrt{3},2)\cup\ldots \end{align*}

We first calculate the limits in the interior of the intervals, then we'll have a look at the left endpoints of $A_n$and $B_n$.

Interior of $A_n$: ${A}^{\circ}_{n}=(\sqrt{n},\sqrt{n+1}), n\geq 0$

Let's fix $n_0\geq 0$ and consider the interior ${A}^{\circ}_{n_0}=(\sqrt{n_0},\sqrt{n_0+1})$. The function $f$ is given by $$f(x)=\frac{n_0}{x^2},\qquad \sqrt{n_0}<x<\sqrt{n_0+1}$$ in the interior of $A_{n_0}$. Let $x_0\in {A}_{n_0}^\circ=(\sqrt{n_0},\sqrt{n_0+1})$, then $$\lim\limits_{x\to x_0^-}\frac{n_0}{x^2}=\lim\limits_{x\to x_0^+}\frac{n_0}{x^2}=\frac{n_0}{x_0^2}$$ Since the left- and right-hand limit of $x_0$ and the function value at $x_0$ coincide, the limit at $x_0$ exists and $$\lim\limits_{x\to x_0}f(x)=\lim\limits_{x\to x_0}\frac{n_0}{x^2}=\frac{n_0}{x_0^2} \qquad\qquad x_0\in {A}_{n_0}^{\circ}, n\geq 0$$

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Interior of $B_n$: ${B}_n^\circ=(-\sqrt{n},-\sqrt{n-1}),n\geq 1$

Let's fix $n_0\geq 1$ and consider the interior ${B}_{n_0}^\circ=(-\sqrt{n_0},-\sqrt{n_0-1})$. The function $f$ is given by $$f(x)=\frac{n_0}{x^2},\qquad -\sqrt{n_0}<x<-\sqrt{n_0+1}$$ in the interior of $B_{n_0}$. Let $x_0\in {B}_{n_0}^\circ=(-\sqrt{n_0},-\sqrt{n_0-1})$, then $$\lim\limits_{x\to x_0^-}\frac{n_0}{x^2}=\lim\limits_{x\to x_0^+}\frac{n_0}{x^2}=\frac{n_0}{x_0^2}$$ Since the left- and right-hand limit of $x_0$ and the function value at $x_0$ coincide, the limit at $x_0$ exists and $$\lim\limits_{x\to x_0}f(x)=\lim\limits_{x\to x_0}\frac{n_0}{x^2}=\frac{n_0}{x_0^2} \qquad\qquad x_0\in {B}_{n_0}^{\circ}, n\geq 1$$

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Left endpoints of $A_n=[\sqrt{n},\sqrt{n+1})$: Points $x_0=\sqrt{n}, n\geq 1$

We exclude the interval $A_0$, since $f$ is not defined at the left endpoint of $A_0=[0,1)$. Let's fix $n_0\geq 1$ and consider the left endpoint $x_0=\sqrt{n_0}$ of $A_{n_0}$. We observe since $A_{n_0-1}=[\sqrt{n_0-1},\sqrt{n_0})$ \begin{align*} \lim\limits_{x\to \sqrt{n_0}^{\,-}}f(x)&=\lim\limits_{x\to \sqrt{n_0}^{\,-}}\frac{n_0-1}{x^2}=\frac{n_0-1}{n_0}=1-\frac{1}{n_0}\\ \lim\limits_{x\to \sqrt{n_0}^{\,+}}f(x)&=\lim\limits_{x\to \sqrt{n_0}^{\,+}}\frac{n_0}{x^2}=\frac{n_0}{n_0}=1\\ f(\sqrt{n_0})&=\frac{n_0}{n_0}=1 \end{align*} Since left- and right-hand limit of $x_0$ do not coincide, the limit at $x_0=\sqrt{n_0}$ does not exist.

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Left endpoints of $B_n=[-\sqrt{n},-\sqrt{n-1})$: Points $x_0=-\sqrt{n}, n\geq 1$

Let's fix $n_0\geq 1$ and consider the left endpoint $x_0=-\sqrt{n_0}$ of $B_{n_0}$. We observe since $B_{n_0+1}=[-\sqrt{n_0+1},-\sqrt{n_0})$ \begin{align*} \lim\limits_{x\to -\sqrt{n_0}^{\,-}}f(x)&=\lim\limits_{x\to -\sqrt{n_0}^{\,-}}\frac{n_0+1}{x^2}=\frac{n_0+1}{n_0}=1+\frac{1}{n_0}\\ \lim\limits_{x\to -\sqrt{n_0}^{\,+}}f(x)&=\lim\limits_{x\to -\sqrt{n_0}^{\,+}}\frac{n_0}{x^2}=\frac{n_0}{n_0}=1\\ f(-\sqrt{n_0})&=\frac{n_0}{n_0}=1 \end{align*} Since left- and right-hand limit of $x_0$ do not coincide, the limit at $x_0=\sqrt{n_0}$ does not exist.

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Summary:

• The function $f(x)=\lim\limits_{x\to x_0}\frac{\lfloor x\rfloor}{x^2}$ has discontinuities exactly at $x\in \mathbb{Z}\backslash\{0\}$

• The limit $\lim\limits_{x\to x_0}\frac{\lfloor x^2\rfloor}{x^2}$ exists for all values $x_0\in \mathbb{R}\backslash\mathbb{Z}$ and $$\lim\limits_{x\to x_0}\frac{\lfloor x^2\rfloor}{x^2}=\frac{\lfloor x_0^2\rfloor}{x_0^2}$$

• The function $f(x)=\frac{\lfloor x^2\rfloor}{x^2}$ has a removable singularity at $x=0$. Since \begin{align*} \lim\limits_{x\to 0^-}\frac{\lfloor x^2\rfloor}{x^2}=\lim\limits_{x\to 0^-}\frac{0}{x^2}=0 \text{ and } \lim\limits_{x\to 0^+}\frac{\lfloor x^2\rfloor}{x^2}=\lim\limits_{x\to 0^+}\frac{0}{x^2}=0 \end{align*} we can define a function \begin{align*} &g:\mathbb{R}\rightarrow\mathbb{R}\\ &g(x)= \begin{cases} f(x)=\frac{\lfloor x^2\rfloor}{x^2}\qquad&x\ne 0\\ 0&x=0 \end{cases} \end{align*} which is continuous at $x=0$.