If for every $\epsilon>0$ there exists $\delta>0$ such that $$|f(x)-f(1)|\geq\epsilon $$ whenever $$|x-1|\geq\delta$$ Then which of the following is true?
A.$f$ is discontinuous at $x=1$
B.$f$ is unbounded.
C.$\int_{0}^{\infty}|f(x)|dx$ exists.
D.$\lim_{|x|\to\infty}|f(x)|=\infty$
I consider an example $f(x)=x$ $\hspace{0.2cm}$ as we go far from $1$ also $f(x)$ goes far from $f(1)$.In this case B,D are true and A,C are false.
The function $f(x) = x$ serves as a counterexample to $A$, by choosing $\epsilon = \delta$. It also serves as a counterexample to $C$.
The function
$$f(x) = \left\{\begin{array}{cc} x :& x \in \mathbb{Z} \\ 0 :& \text{ else } \end{array}\right.$$
serves as a counterexample to $D$. Note how $f(x)$ gets arbitrarily far away from $f(1)$, but is small for almost all values of $x$.