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I have a question,we know that limit of $3x$ when $x$ approaches $2$ is $6.$ We can prove that with epsilon and delta definition. Suppose I wrongly assume the limit is $7$ then by definition for some epsilon I must be unable to find a delta. What is that epsilon? Just trying understand the notion of limit using counter example. That is for what value of epsilon this fails? $|3x-7|< \epsilon$ whenever $|x-2|< \delta.$

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    $\begingroup$ Why didn't you mark any of the answers as correct? $\endgroup$ Commented Jan 23, 2020 at 12:13
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    $\begingroup$ Can I do that? I thought only those who have a higher reputation can do it. Never knew this. $\endgroup$
    – danny
    Commented Jan 23, 2020 at 12:15
  • $\begingroup$ No idea if you're trolling or serious, but it's disrespectful for people to use their time to answer questions without getting anything from it $\endgroup$ Commented Jan 23, 2020 at 12:17
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    $\begingroup$ No I’m seriously not trolling. I thought upvoting and marking answers as correct can only be done by people with high reputation. Thank you for letting me know. $\endgroup$
    – danny
    Commented Jan 23, 2020 at 12:19

3 Answers 3

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In a more general setting: suppose that $\lim_{x\to a}f(x)=b$ and $c\neq b$.

Then $|b-c|>0$ so we can take $\epsilon:=\frac12|b-c|$.

The triangular inequality tells us that: $$2\epsilon=|b-c|\leq|f(x)-b|+|f(x)-c|$$

So whenever we have $|f(x)-b|<\epsilon$ we also must have $|f(x)-c|>\epsilon$.

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Let epsilon = 1, so for any x such that $|x-2|<\delta$ then $|3x-7|<1$
Since $|2-2+\frac{\delta}{2}|< \delta $ it should $|3\cdot (2-\frac{\delta}{2})-7|<1$ or $|-1-\frac{3\delta}{2}|<1$ since $\delta>0$ then we get $1+\frac{3\delta}{2}<1$ or $\frac{3\delta}{2}<0$ so contradiction.

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I notice that two answers have offered two distinct values of $\epsilon.$ This reflects the fact that there is no such thing as "the" value of $\epsilon$ that you should use in a counterexample; there are many values of epsilon that you can use.

In the particular example you mention, whenever $x < 2,$ then $\lvert 3x - 7 \rvert = 7 - 3x > 1.$

Now let $0 < \epsilon \leq 1,$ that is choose any value of $\epsilon$ you like as long as it is positive and not greater than $1.$ Then there is no positive value of $\delta$ such that $\lvert 3x - 7 \rvert < \epsilon$ whenever $\lvert x-2 \rvert < \delta,$ because no matter what positive value of $\delta$ we might try, the value $x = 2 - \frac\delta2$ satisfies $\lvert x-2 \rvert < \delta$ but not $\lvert 3x - 7 \rvert < \epsilon.$

Personally, unless there is a need to identify exactly when a statement is true and when it is false, I like counterexamples that are not "right on the edge," so in your particular example I'd probably set $\epsilon = \frac12$ rather than $\epsilon = 1.$ (Why $\frac12$? Because I want a number between zero and $1,$ so I might just split the difference.) But that's just personal preference. You can set $\epsilon = 1,$ or you can set $\epsilon = \frac{1}{1000}.$

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