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!
Best Answer
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 $(x-1, x]$.