I'm reading the book "Calculus a Rigorous First Course" by Daniel J Velleman. Currently stuck on exercise 11 from section 2.3 which is the following.
Use the given formula for $\delta$ to prove the limit statement.
\begin{align*} \lim_{x\rightarrow3}\lfloor x/2 \rfloor=1 \; ,\; \delta=1 \end{align*}
My attempt was the following.
Proof. Suppose $\epsilon>0$. Let $\delta = 1$. Then
\begin{align*} | \lfloor x/2 \rfloor - 1| < 1 = \delta=\epsilon \end{align*}
Note, $| \lfloor x/2 \rfloor - 1| < 1$ because $2<x<4$.
I find this proof somewhat incorrect because $\delta \neq \epsilon$. Or am I missing something about the epsilon-delta criterion?
You're right that $\delta \ne \epsilon$, so that doesn't belong in the proof. Also, you have skipped some steps in your writeup of the proof. In particular, you never said that you are assuming that $0 < |x - 3| < \delta$. Here's how I would write it:
Suppose $\epsilon > 0$. Let $\delta = 1$. Suppose $0 < |x - 3| < \delta$. Then $|x - 3| < 1$, so $2 < x < 4$. Therefore $1 < x/2 < 2$, so $\lfloor x/2 \rfloor = 1$ and $|\lfloor x/2 \rfloor - 1| = 0 < \epsilon$.