Given the following statement:
There is $\epsilon > 0$ and $\delta > 0$ such that $|F(x) - \pi| < \epsilon$ whenever $0 < |x-4| < \delta$.
Provide an example of a function F for which the statement is true, yet $\lim\limits_{x \to 4}f(x) = \pi$ fails.
I think that this statement being false for $\lim\limits_{x \to 4}f(x) = \pi$ has something to do with the fact that it implies that there "exists" an $\epsilon$ rather than saying for "all" $\epsilon$, but am unclear where to go from there.
Could someone please guide me into the correct direction?
For example for $\varepsilon=3$
$$F(x)=\pi+\frac{x-4}\delta-1$$
then
$$\left|F(x) - \pi\right| =\left|\frac{x-4}\delta -1\right|< \varepsilon$$
is true but $F(x) \to \pi-1$.
More in general, for any fixed $\varepsilon>0$ let consider
$$F(x)=\pi+\frac \varepsilon 2\frac{x-4}\delta+\frac \varepsilon 2$$
then
$$\left|F(x) - \pi\right| =\left|\frac \varepsilon 2\frac{x-4}\delta+\frac \varepsilon 2\right|< \varepsilon$$
is true but $F(x) \to \pi+\frac \varepsilon 2$.