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If $f(x)$ is discontinuous at 0, does that mean $f(x) + \frac{1}{f(x)}$ is also discontinuous?

I know that if $f(x)$ and $g(x)$ are both discontinuous at 0, that does not necessarily mean that $f(x) + g(x)$ is discontinuous at 0, (take $f(x) = \frac{1}{x}$ and $g(x) = x - \frac{1}{x}$), but I cannot think of an example to show $f(x) + \frac{1}{f(x)}$ is continuous even if $f(x)$ is discontinuous.

Does this mean that it is impossible?

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  • $\begingroup$ It might be interesting that the answer is different if you consider $f - 1/f $ (assuming $f>0$) instead of $f + 1/f $. $\endgroup$ – PhoemueX Apr 5 '16 at 15:59
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    $\begingroup$ Actually, $x\to1/x$ is not discontinuous at $0$, it is undefined. This function happens to be continuous everywhere on its domain of definition. Common mistake. $\endgroup$ – Jean-Claude Arbaut Apr 5 '16 at 16:05
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Consider the function: $$f(x)=\begin{cases}2 &\text{ if }x\ne0\\\frac12&\text{ if }x=0\end{cases}$$

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Consider $$ f (x) = \begin {cases} 2 , &\text{ if } x \in \Bbb {Q},\\ 1/2, &\text { if } x \notin \Bbb {Q}. \end {cases} $$ Then $f + 1/f \equiv 5/2$ is continuous, but $f $ is not.

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    $\begingroup$ Yes, but 5/2 not 3/2 $\endgroup$ – almagest Apr 5 '16 at 16:00
  • $\begingroup$ @almagest: Whoops. Of course. Thank you. $\endgroup$ – PhoemueX Apr 5 '16 at 16:56
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If you allow complex-valued functions, it is easy to show that the answer is no: consider $$ f(x) = \begin{cases} i & x\leq 0 \\ -i & x > 0 \end{cases} $$ Obviously $f(x)$ is discontinuous at $x=0$. However, if $x \leq 0$, $$ f(x)+\frac{1}{f(x)} = i+\frac{1}{i} = 0, $$ and similarly, $ -i+1/(-i)=0$. Hence $$ f(x) + \frac{1}{f(x)} = 0 $$ for any real $x$, which is obviously continuous.

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