Is the "limit function" a continuous function? 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?
 A: 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
$$|g(x)-g(x_0)|<|g(x)-f(x')|+|f(x')-g(x_0)|<{\epsilon\over2}+{\epsilon\over2}=\epsilon\ .$$
A: 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.
