- #1
razored
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Prove that the limit exists :[tex]\lim_{x\rightarrow 1} 3x -1 =2[/tex]
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.
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.