If $f$ is a real function and $g(x)=\lim_{t\to x}f(t)$, then $g$ is continuous? Suppose that a real function $f$ has a limit at every point in a set $K\subset\mathbb{R}$ and $$g(x)=\lim_{t\to x}f(t)$$ Does that imply that $g$ is continuous on $K$?
 A: According to the assumption we have limits 
i) $g(x) = \lim_{t\to x} f(t)$ and
ii) $g(x+h) = \lim_{t\to x+h} f(t)$
What we want is to have $g(x) = \lim_{h\to 0} g(x+h)$
That is $\forall\epsilon$ $\exists\delta$ s.t. if $|h|<\delta$ then $|g(x+h)-g(x)|<\epsilon$.
Choose $\epsilon=\epsilon_0$ 
By (i) for the $\epsilon=\epsilon_0/2$ $\exists\delta_0$ s.t. $|t-x|<\delta_0$ and $|f(t)-g(x)|<\epsilon_0/2$
By (ii) for the $\epsilon=\epsilon_0/2$ $\exists\delta_1$ s.t. $|t-x-h|<\delta_1$ and $|f(t)-g(x+h)|<\epsilon_0/2$
Then picking our $\delta = \delta_0+\delta_1$ we have
$|t-x|<\delta_0$ & $|t-x-h|<\delta_1$ implies $|h|<\delta_0+\delta_1$
with 
$|f(t)-g(x)|<\epsilon_0/2$ & $|f(t)-g(x+h)|<\epsilon_0/2$ implies $|g(x+h)-g(x)|<\epsilon_0$
To sum up, given $\epsilon$ we can choose $\delta$ as above s.t. limit conditions is satisfied. Therefore we say g is continuous.
A: $g$ does need to be continuous, provided $K$ does not contain any isolated points. Let $x_0$ be in $K$. Let $g(x_0) = L$. Let $ϵ>0$. Then, by the definition of $g$, there exists $δ>0$ such that for all $x$ in $K$ with $0<|x-x_0|<δ$, $|f(x)-L|<ϵ/2$. But for each $y$ in $K$ with $0<|y-x_0|<δ$, we also have $|f(x)- g(y)|<ϵ/2$ for $x$ in $K$ and with $0<|x-x_0|<δ$ sufficiently close to $y$. Therefore, for each $y$ in $K$ with $0<|y-x_0|<δ$, choosing such an $x$, $|g(y)-g(x_0)|=|g(y)-L|=|-f(x)+g(y)+f(x)-L|≤ |f(x)-L| + |f(x)-g(y)|<ϵ/2 + ϵ/2=ϵ$, so $g$ is continuous on $K$. However, if there are allowed to be points in $K$ with no other points in some interval containing them (isolated points), then a counterexample exists. Let $K$ be the set consisting of $0$ and $1/n$ for each $n$ a positive integer, and let $f(x)=x$. $g$ can be defined however one wishes on $1/n$, say $g(1/n)=1$, as the limit properties will hold vacuously, but we must have $g(0)=0$, in which case $g$ is discontinuous at $0$. 
A: For a function to be continuous on an interval, 3 conditions must be met:


*

*f(t) must be defined.

*The limit must exist.

*The limit equals f(t).


What does this say of g?
