If I have a function $f(x)$ defined as follows.
- $f(x) = 1$ for all $x<1$ and $x>2$;
- $f(x) = 100$ for $x = 1.5$;
- $f(x)$ is undefined anywhere else.
According to the $\varepsilon$–$\delta$ definition of continuity, if I take $\delta$ as any positive number smaller than $0.5$, then $f(x)$ by definition is continuous at $x = 1.5$ because within the $\delta$-neighborhood there is only one point defined, but $f(x)$ is obviously not continuous at $x = 1.5$.
Below is the $\varepsilon$–$\delta$ definition of continuity:
The function $f(x)$ is continuous at a point $x_0$ of its domain if for every positive $\varepsilon$ we can find a positive number $\delta$ such that $$|f(x) – f(x_0)|<\varepsilon$$ for all values $x$ in the domain of $f$ for which $|x-x_0|<\delta$.
Best Answer
Your example in fact shows that according to the formal definition of continuity, the function $f$ as you have defined it is continuous at $x=1.5$, and rather your informal suggestion that $f$ is "obviously not continuous at $x=1.5$" is actually mistaken.