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Consider the function $$f(x) = \frac{x^3}{1+x^3}$$enter image description here

Obviously this function is discontinuous at $x = -1$ therefore discontinuous on $\mathbb{R}$. Moreover, it is unbounded at the same point. Now, I would not say that this function is locally bounded either, as not all sets $f(A)$ are bounded for a neighbourhood $A$ about any $x_0 \in \mathbb{R}$. Is this reasoning correct? Can an unbounded discontinuous function be locally bounded?

mesllo
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  • Does locally bounded require the $x_{0}\in \mathbb{R}$? or $x_{0}$ in the domain of $f(x)$? – TravisJ Apr 22 '15 at 13:14
  • @TravisJ The question pretty clearly says $x_0\in\mathbb R$... – 5xum Apr 22 '15 at 13:15
  • I conjecture that it would be more plausible to guess that every function of discontinuity of infinity is not locally bounded. – Yes Apr 22 '15 at 13:15
  • @5xum, I know it says $\mathbb{R}$, but I was curious if that was a mistake. – TravisJ Apr 22 '15 at 13:16
  • @chou What is "a function of discontinuity of infinity"? – 5xum Apr 22 '15 at 13:17
  • @TravisJ That's an issue for me personally. I am a kind of unsure over how the domain is considered. If that is the case, this function would be locally bounded on the domain $[0,\infty)$ if $x_0$ is only considered within that domain. Then again it would be continuous over it as well which goes against my question. – mesllo Apr 22 '15 at 13:18
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    Well the question is what function would not be locally bounded if we argued for having $x$ be only confined to the domain of the function. Essentially functions that would not be locally bounded under that definition would have to be discontinuous. Since they would need to be defined at points $a$ such that $\lim_{x\to a}|f(x)|=\infty$. – DRF Apr 22 '15 at 13:39
  • This leads me to ask: does it really make sense to analyse functions on one particular domain, by using an $x_0$ NOT in that domain? – mesllo Apr 22 '15 at 13:45
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    @jablesauce It does, that's what poles are. It makes sense to say that $1/x$ has a pole (of order $1$) at $0$, even though $0$ is not in the domain of $1/x$. – 5xum Apr 22 '15 at 13:58
  • Aa yes of course! – mesllo Apr 22 '15 at 14:00
  • @5xum: Yeah, abuse of language; I meant "infinite discontinuity" – Yes Apr 22 '15 at 14:13

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If I understand your definition correctly, the function $$f(x) = \begin{cases}x - 1 & x < 0\\ 0 & x = 0\\ x+1 & x>0\end{cases}$$

fits the bill.

If you want a rational function, then no can do, because if $p, q$ are polynomials with no common factors, then $f(x)=\frac{p(x)}{q(x)}$ is continuous if $q(x)\neq 0$ and is unbounded around $x$ if $q(x)=0$.

5xum
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