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 :)


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 $x_0$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.