Counter example for limit 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 $|x-2|< \delta.$
 A: 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. 
A: 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 $|f(x)-c|>\epsilon$.
A: 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 $\epsilon = \frac{1}{1000}.$
