3
$\begingroup$

Does $\lim_{x \to 0+} \left(x\lfloor \frac{a}{x} \rfloor\right)=a?$

I'm going to say this statement is false, and try to use the properties of limits

$$\lim_{x \to 0+} \left(x\lfloor \frac{a}{x} \rfloor\right)=\lim_{x \to 0+} x.\lim_{x \to 0+}\lfloor \frac{a}{x}\rfloor$$ =$0.\infty$ which is undefined.

Or is it $\infty $ because x ends up being a little bigger than 0?

$\endgroup$
3
  • $\begingroup$ Situations where you have e. g. "zero times infinity" are complicated, because you have to either manipulate the expression to something else that can be explicitly solved (or found to have no solution), or make an argument about which of the two happens faster. $\endgroup$ – MathInferno Jul 30 '16 at 15:26
  • 1
    $\begingroup$ You can only split the limit of the product if each limit exists and is finite $\endgroup$ – Aweygan Jul 30 '16 at 15:27
  • $\begingroup$ @Aweygan you're right, thanks for the reminder. $\endgroup$ – RonaldB Jul 30 '16 at 15:28
4
$\begingroup$

Hint: Suppose that $a$ is positive. Then $$\left\lfloor\frac{a}{x}\right\rfloor=\frac{a}{x}-\delta(a,x),$$ where $0\le \delta(a,x)\lt 1$. Now multiply by $x$ and take the limit.

The case $a$ negative will be very similar, and $a=0$ is obvious.

$\endgroup$
2
  • $\begingroup$ Thank you I believe i got it. So the statement holds true. $\endgroup$ – RonaldB Jul 30 '16 at 15:32
  • 1
    $\begingroup$ Good! Yes, the statement holds true. There is no need to separate cases. However, your geometric intuition is likely to be better for positive $a$ (mine certainly is) so I wanted to start with that. $\endgroup$ – André Nicolas Jul 30 '16 at 15:35
4
$\begingroup$

$$\frac{a}{x} - 1 \le E(\frac{a}{x}) \le \frac{a}{x}$$

$$a- x \le xE(\frac{a}{x}) \le a$$

if you apply the limit it gives $a$

Note: $E(x)$ is the step function.

$\endgroup$
1
  • $\begingroup$ No it does not. The upper bound and lower bound do not meet in the limit, the squeeze theorem does not apply here. $\endgroup$ – Jack Tiger Lam Jul 30 '16 at 15:44
1
$\begingroup$

$\bullet$ First show $\displaystyle \lim_{x\to \infty}\frac{[x]}{x}=1$. For this ,

$$\left|\frac{[x]}{x}-1\right|=\left|\frac{[x]-x}{x}\right|<\frac{1}{|x|}\to 0 \text{ as , } x\to \infty.$$

$\bullet$ Now put , $\frac{a}{x}=y$. Then limit becomes $\displaystyle \lim_{y\to \infty}\frac{a[y]}{y}=a$ .

$\endgroup$
5
  • $\begingroup$ How $\frac{a}{x}=y$ imply $\displaystyle \frac{[x]}{x}=\frac{a[y]}{y}$ $\endgroup$ – user1942348 Mar 26 '17 at 14:56
  • $\begingroup$ @user1942348 What a silly question..!! Put , $a/x=y$ in the limit given in question. $\endgroup$ – Empty Mar 26 '17 at 20:15
  • $\begingroup$ $a/x=y$ implies as $x\to \infty$ then $y \to 0$. Also $\displaystyle \frac{[x]}{x}=\frac{y[a/y]}{a}=?\frac{a[y]}{y}$ $\endgroup$ – user1942348 Mar 27 '17 at 4:49
  • $\begingroup$ @user1942348 First read the question carefully..It is given that $x\to 0^+$. So from where you got $x\to \infty$ ? $\endgroup$ – Empty Mar 27 '17 at 6:53
  • $\begingroup$ I've told you that put $a/x=y$ in the question..NOT in my answer. Do you know what the QUESTION means ? $\endgroup$ – Empty Mar 27 '17 at 6:54

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.