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If $\lim_{x\to 1} f(x)$ and $\lim_{x\to 1} g(x)$ do not exist, does the

$\lim_{x\to 1}{ (f+g)}$, exist? If so, how?

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closed as off-topic by Nosrati, Gibbs, heropup, kimchi lover, Adrian Keister Sep 7 '18 at 0:11

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    $\begingroup$ $f(x) = -g(x) = 1/(x-1)$ would do. $\endgroup$ – xbh Sep 5 '18 at 15:22
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$0=f-f$ so even if $lim_{x\to1}f$ may not exist....$lim_{x\to1}f+g$ can exist...

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Here's another suggestion using the Dirichlet function, take: $$f(x) = \begin{cases} 0 & x \in \mathbb{R}\setminus\mathbb{Q}\\ 1 & x \in \mathbb{Q} \end{cases}$$ and $$g(x) = \begin{cases} 1 & x \in \mathbb{R}\setminus\mathbb{Q}\\ 0 & x \in \mathbb{Q} \end{cases}$$ Both are nowhere continuous, so $\lim_{x \to 1}f(x)$ and $\lim_{x \to 1}g(x)$ definitely don't exist.

However, for the sum you simply have $f(x)+g(x)=\ldots$, so ...

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