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Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that for all $x_0\in\mathbb{R}$ we have $\lim\limits_{x\to x_0}f(x)=g(x_0)\in \mathbb{R}$.

Is $g$ a continuous function?

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Have you researched on the web for Continous functions. Here is a good link: en.wikipedia.org/wiki/Continuous_function –  Maxood Mar 16 '12 at 15:49
What have you tried? –  Chris Eagle Mar 16 '12 at 15:51
By considering the constant sequence you directly have $f=g$ and then you have the definition of continuous –  Listing Mar 16 '12 at 15:59
@Listing: Constant sequences (and, more generally, sequences $(x_n')$ such that $x_n' = x_0$ for some $n$) are likely to be excluded. There was some discussion about this definition some time ago, although I don't remember in which topic. Hence, the arguments one has to use are more sophisticated, and $f$ may no be continuous. –  D. Thomine Mar 16 '12 at 16:09
See ALso: math.stackexchange.com/questions/3777/… –  Aryabhata Mar 16 '12 at 17:49
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2 Answers

Fix an $x_0$ and let an $\epsilon>0$ be given. We have to prove that there is a $\delta>0$ such that $|g(x)-g(x_0)|<\epsilon$ as soon as $|x-x_0|<\delta$.

By assumption on $f$ and $g$ there is a $\delta>0$ such that $$|f(x)-g(x_0)|<{\epsilon\over2}\qquad \bigl(0<|x-x_0|<\delta\bigr)\ .$$ Consider now an arbitrary $x$ with $|x-x_0|<\delta$. There is a $\delta'>0$ such that $$|f(x')-g(x)|<{\epsilon\over2}\qquad \bigl(0<|x'-x|<\delta'\bigr)\ .$$ The open set $$S:=\ ]x_0-\delta, x_0+\delta[\ \cap\ ]x-\delta',x+\delta'[\ \ \setminus\{x_0,x\}$$ is nonempty; therefore we may take a point $x'\in S$ and then have $$|g(x)-g(x_0)|<|g(x)-f(x')|+|f(x')-g(x_0)|<{\epsilon\over2}+{\epsilon\over2}=\epsilon\ .$$

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Take $\varepsilon>0$.

By definition of the limit (assuming the limit in the statement is well defined, otherwise the question has no meaning) there exists some $\delta>0$ such that on the punctured interval $(x_0-\delta,x_0+\delta)\setminus\{x_0\}$ we get $|f(x)-g(x_0)|<\varepsilon/2$. In particular every $g(x)$ on this interval has to be strictly within $\varepsilon$ of $g(x_0)$, since values of $f$ are bounded to within $\varepsilon/2$ of $g(x_0)$ so any limit on values of $f$ (which define $g$) will also subsequently be bounded to within $\varepsilon$, which proves continuity of $g$.

Note that this isn't based on the continuity of $f$, which is not required for $g$ to be continuous, for example $f$ can contain infinitely many isolated removable discontinuities, and $g$ is still happily continuous.

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A concrete example of a discontinuous $f$ is Thomae's function -- for which $g$ is identically zero. –  Henning Makholm Mar 16 '12 at 16:27
There is something missing here. The statement "In particular every $g(x)$ on this interval has to be strictly within $\varepsilon$ of $g(x_0)\ $" needs clarification. –  Christian Blatter Mar 16 '12 at 20:33
@ChristianBlatter, how's that for clarification? –  davin Mar 16 '12 at 20:45
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