Prove the statement using the ε, δ definition of a limit. lim x → 1 4 + 2x 3 = 2

Given ε > 0, we need δ ---Select--- such that if 0 < |x − 3| < δ, then 2 + 1 3 x − 3 ---Select--- . But 2 + 1 3 x − 3 < ε ⇔ 1 3 x − 1 < ε ⇔ 1 3 |x − 3| < ε ⇔ |x − 3| < ---Select--- . So if we choose δ = ---Select--- then 0 < |x − 3| < δ ⇒ 2 + 1 3 x − 3 < ε. Thus, lim x → 3 2 + 1 3 x = 3 by the definition of a limit.

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LammettHash LammettHash · Answer 1 · 2020-09-15T19:52:33+02:00

You want to prove

[tex]\displaystyle\lim_{x\to1}\frac{4+2x}3=2[/tex]

By the definition of the limit, this means that for any [tex]\varepsilon>0[/tex], we can find [tex]\delta>0[/tex] such that

[tex]|x-1|<\delta\implies\left|\dfrac{4+2x}3-2\right|<\varepsilon[/tex].

In order to get the inequality we want, we need to have

[tex]\left|\dfrac{4+2x}3-2\right|=\left|\dfrac{2x-2}3\right|=\dfrac23|x-1|<\varepsilon\implies|x-1|<\dfrac{3\varepsilon}2[/tex]

so we can pick [tex]\delta=\frac{3\varepsilon}2[/tex].

So here's the proof: Let [tex]\varepsilon>0[/tex] be given, and let [tex]\delta=\frac{3\varepsilon}2[/tex]. Then

[tex]|x-1|<\delta=\dfrac{3\varepsilon}2[/tex]

[tex]\implies\dfrac23|x-1|<\varepsilon[/tex]

[tex]\implies\left|\dfrac{2x-2}3\right|=\left|\dfrac{2x+4}3-1\right|<\varepsilon[/tex]

as required.