Counter example for limit

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 $|f(x)-c|>\epsilon$.

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.