Given that lim x→2 (4x − 2) = 6, illustrate Definition 2 by finding values of δ that correspond to ε = 0.5, ε = 0.1, and ε = 0.05.ε = 0.5 δ ≤ ?ε = 0.1 δ ≤ ?ε = 0.05 δ ≤ ?

Respuesta :

More generally, let [tex]\delta=\frac\varepsilon4[/tex]. Then

[tex]|x-2|<\delta=\dfrac\varepsilon4\implies4|x-2|=|4x-8|=|(4x-2)-6)|<\varepsilon[/tex]

which proves the limit is 6. So

[tex]\varepsilon=0.5\implies\delta=0.125[/tex]

[tex]\varepsilon=0.1\implies\delta=0.025[/tex]

[tex]\varepsilon=0.05\implies\delta=0.0125[/tex]

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