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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!

  • 1
    What happens when you take $5-\varepsilon < 2x+3 < 5+\varepsilon$ and subtract though by $3$ and then divide through by $2$? – 2017-01-23

1 Answers 1

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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 $\delta = \epsilon/ 2$