Consider the following typographical error in the limit definition: $$\forall \epsilon \gt 0, \exists \delta \gt 0:x \in X, 0 \lt \lvert x-a \rvert \lt \epsilon\Rightarrow \lvert f(x) – L \rvert \lt \delta$$ Show that $f$ satisfies this condition if and only if it is bounded on any bounded interval from center to. In the affirmative case, L can be any real number.
My answer
(→) Let $I$ be an interval of length $\epsilon$ and centered on $a$. Then $I \subset (a-\epsilon, a+\epsilon)$ and by the hypothesis, there is $\delta \gt 0$ such that $\lvert f(x)\rvert -\lvert L \rvert \le \lvert f (x) – L \rvert \lt \delta$ for all $x \in (a-\epsilon, a+\epsilon)$. So, for every $x \in I$ we have that $\lvert f(x)\rvert \lt \delta + \lvert L \rvert$ and consequently, $f$ is bounded on I.
(←) Let $\epsilon \gt 0$, There is $A \in \mathbb R$ such that for all $x \in (a-\epsilon, a+\epsilon)$ we have $\lvert f(x) \rvert \lt A$. Hence we have that for all $x\in \mathbb R$ such that $\lvert x-a \rvert \lt\epsilon$ we have $\lvert f(x)-L \rvert \le \lvert f(x) \rvert + \lvert L \rvert \lt A + \lvert L \rvert$. Thus, taking $\delta = A + \lvert L \rvert$ we will have the condition we wanted and as $\epsilon$ is arbitrary the statement is proved.
It is acceptable?
Thanks for any help.