I don't understand how to solve this problem and the official solution does not make much sense to me either.
The problem is:
(d) Let $f$ be a function with the property that every point of discontinuity is a removable discontinuity. This means that $\lim_{y\to x}f(y)$ exists for all $x$, but $f$ may be discontinous at some (even infinetely many) numbers $x$. Define $g(x) = \lim_{y\to x}f(y)$. Prove that $g$ is continous.
The official solution is:
Since $g(a) = \lim_{y\to a}f(y)$, by definition, it follows that for any $\varepsilon>0$ there is a $\delta>0$ such that $|f(y)-g(a)|<\varepsilon$ for $|y-a|<\delta$. This means that $$g(a)-\varepsilon<f(y)<g(a)+\varepsilon$$ for $|y-a|<\delta$. So if $|x-a|<\delta$, we have $$g(a)-\varepsilon \leq \lim_{y\to x}f(y) \leq g(a)+\varepsilon$$ which shows that $|g(x)-g(a)|\leq\varepsilon$ for all $x$ satisfying $|x-a|<\delta$. Thus $g$ is continous at $a$.
I don't understand how saying that $|x-a|<\delta$ impliest that $g(a)-\varepsilon \leq \lim_{y\to x}f(y) \leq g(a)+\varepsilon$.
When I was trying to come up with my own solution the thing that bothered me that if a lot of values of $f(x)$ are changed (so that they are equal to $\lim_{x\to a}f(x)$) then this may in fact change the limit of some other points? I then tried to say that for some $\delta>0$ the values for $x$ in $0<|x-a|<\delta$ the values would stay the same, however I think that for a functions such as one where $f(x) = 0$ if $x$ is irrational and $f(x) = \frac{1}{q}$ if $x=\frac{p}{q}$ in lowest terms, this would not be true. Where did I go wrong?
