Question

prove using epsilon and delta that the limit of [x/2] as x approaches 3 =1

Answer 1

\[\lim_{x \rightarrow 3} \lfloor x/2 \rfloor=1\]

Answer 2

\[
\left\lfloor
\frac{x}{2}\right\rfloor =1, \text { for } 1\le x \le 4
\]
near 3 it is equal to 1
Let \( \epsilon >0\), take \(\delta =\frac 1 2\), then if
\[| x -3| < \delta \implies \left | \left\lfloor
\frac{x}{2}\right\rfloor -1 \right |=|1-1|=0 <\epsilon
\]

Answer 3

"let delta be a 1/2 " that was randomly picked number yes?
Second question : how did you get to x-3?

Answer 4

I think it's for \[1\le x <4\] and not including 4. Then I understand.

Answer 5

yes, you are right.