Prove that the limit exists :[tex]\lim_{x\rightarrow 1} 3x -1 =2[/tex](adsbygoogle = window.adsbygoogle || []).push({});

Here is the work to find [tex]\delta[/tex].

[tex]|3x-3|<\epsilon[/tex] Substitution

[tex]3|x-1|<\epsilon[/tex]

[tex]|x-1|<\frac{\epsilon}{3}[/tex]

[tex]0<|x-1|<\delta [/tex] Appears that epsilon is equal to delta.

[tex]\delta = \frac{\epsilon}{3}[/tex]

I have gotten here thus far,now what?:uhh: What I have proved? The only thing I have discovered that delta is simply equal to epsilon.

I don't understand.

I also understand that you begin the proof like this:

Given [tex]\epsilon> 0[/tex], choose [tex]\delta = \frac_{\epsilon}{3}[/tex]

Now what from here as well?

Thanks beforehand.

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# Homework Help: Epsilon-Delta Proof

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