I'm struggling to prove the following theorem:
For every $ε>0$ there exists a $δ>0$ such that:
$$1-δ < x < 1+δ \implies 5-ε<2x+3<5+ε$$
My thoughts so far are that we want to prove that no matter how small ε is, there is a corresponding δ. I have no experience with epsilon/delta proofs so far, so I would appreciate any advice on how I should proceed. Thanks!
Well, since $\epsilon > 0 $ is given, we have
$$ 5 - \epsilon < 2x + 3 < 5 + \epsilon \iff - \epsilon < 2(x-1) < \epsilon \iff - \epsilon/2 < x-1 < \epsilon/2$$
Thus, if we want $\delta > 0$ to satisfy
$$ 1 - \delta < x < 1 + \delta$$
which is equivalent to
$$ - \delta < x - 1 < \delta $$
then, we better choose $\delta = \epsilon/ 2$