5
$\begingroup$

I have to provide a proof for a function but I'm struggling to grasp the main concept.

$$\lim_{x \to 3} \left\lfloor \frac{x}{2}\right\rfloor = 1 $$

Here is what I've come up with: $$\frac{x}{2} - 1 \lt \left\lfloor \frac{x}{2}\right\rfloor \lt \frac{x}{2} + 1$$

But I'm stuck from here since I cannot use anything else but $\varepsilon$-$\delta$ (meaning no squeeze theorem.) Each of the limits (for every side) is not helpful. I get: $$ \frac{1}{2} \lt \lim_{x \to 3} \left\lfloor \frac{x}{2}\right\rfloor \lt \frac{5}{2}$$

Any clarifications are welcome! Thanks!

$\endgroup$
0

2 Answers 2

5
$\begingroup$

Hint: First, note that $\lfloor x \rfloor$ is always an integer. Use this fact along with the fact that $x -1 < \lfloor x \rfloor \leq x$.

EDIT: By the way, your goal is to show that for all $\epsilon > 0$, there exists a $\delta > 0$, such that

$$0 < |x - 3 | < \delta \Longrightarrow \left| \left\lfloor \frac{x}{2} \right\rfloor - 3\right| < \epsilon.$$

This should be doable with the two previous observations. Your observation you mention is correct, but is not helpful (for your proof, anyways) because your interval $(1/2, 5/2)$ includes two integer values. On the other hand, the inequalities $x - 1 < \lfloor x \rfloor \leq x$ show that $\lfloor x \rfloor$ is the unique integer in the interval $(x-1, x]$.

$\endgroup$
1
  • $\begingroup$ After developing $0 < |x - 3| < \delta $ I got $\frac{1}{2} < \frac{x}{2} < 2$ which is ok if I choose $\delta = \frac{1}{2}\epsilon$ $\endgroup$
    – Cu7l4ss
    Commented Apr 11, 2011 at 7:15
3
$\begingroup$

You can just use the fact that $\lfloor \frac{x}{2}\rfloor $ is constant on $[2,4)$, so for any $\varepsilon >0$, $\delta =0.5 $(or any convenient number $<1$) works.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .