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I was given the following function:

$f(x)=\frac{\left\lfloor x^{2}\right\rfloor }{x^{2}}$

And I need to find the limit of this function for an arbitrary $\underset{x\rightarrow x_{0}^{+}}{\lim}\frac{\left\lfloor x^{2}\right\rfloor }{x^{2}}$

as well as $\underset{x\rightarrow x_{0}^{-}}{\lim}\frac{\left\lfloor x^{2}\right\rfloor }{x^{2}}$

so far I could invent only this :

$\underset{x\rightarrow x_{0}^{+}}{\lim}x^{2}=x_{0}^{2}$

$\underset{y\rightarrow y_{0}^{+}}{\lim}\left\lfloor y\right\rfloor =\max\{m\in\mathbb{Z}|m\leq y\}$

$\underset{x\rightarrow x_{0}^{+}}{\lim}\left\lfloor x^{2}\right\rfloor =\max\{m\in\mathbb{Z}|m\leq x_{0}^{2}\}$

$\underset{x\rightarrow x_{0}^{+}}{\lim}f(x)=\frac{\max\{m\in\mathbb{Z}|m\leq x_{0}^{2}\}}{x_{0}^{2}}$

I have no idea how to proceed or approach this question. I could use some help. Thanks :)

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HINT: You’ll need to consider several cases. For example, suppose that $x_0^2$ is not an integer. In that case you should have no trouble verifying that

$$\lim_{x\to x_0^+}\frac{\left\lfloor x^2\right\rfloor}{x^2}=\frac{\left\lfloor x_0^2\right\rfloor}{x_0^2}=\lim_{x\to x_0^-}\frac{\left\lfloor x^2\right\rfloor}{x^2}\;.$$

Things are a bit more complicated when $x_0^2$ is an integer:

  • the limits from the left and right are not equal, and
  • you’ll need to distinguish the cases $x_0<0$, $x_0=0$, and $x_0>0$.

Note that in these cases the value of $\left\lfloor x^2\right\rfloor$ depends on whether $x^2\ge x_0^2$ or $x^2<x_0^2$, even when $x$ is very close to $x_0$.

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  • $\begingroup$ Just to verify : 1st case - $x_0\notin \mathbb{Z}$ second case is $x_0<0,x_0>0,x_0=0$ and this should be done for both right and left limits ? regarding the last part of what you said - I saw the graph in wolframalpha and when x between 1 and -1 the graph is a zero flatline. how does it makes sense ? $\endgroup$ – Pavel Penshin Mar 29 '16 at 17:59
  • $\begingroup$ @user313448: No, the first case is that $x_0^{\color{red}2}$ is not an integer. For instance, $x_0=\sqrt2$ is not in this case, because its square is the integer $2$. Yes, when $x_0^2$ is an integer you’ll have to distinguish the subcases $x_0<0$, $x_0=0$, and $x_0>0$. When $0<|x|<1$ we have $0<x^2<1$, so $\left\lfloor x^2\right\rfloor=0$, and the fraction is $0$. $\endgroup$ – Brian M. Scott Mar 29 '16 at 18:07

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