Continuous limit function

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function such that the limit $\lim \limits_{y \rightarrow x} f(y)$ exists (and is in fact a real number) for all $x \in \mathbb{R}$.Let us call this limit $g(x)$. Prove that $g$ is continuous throughout $\mathbb{R}$.

Solution

Let $x_0 \in \mathbb{R}$ and $\epsilon>0$. All we have to prove is that there exists a $\delta>0$ such that for $x \in (x_0-\delta, x_0+\delta)$

$$|g(x)-g(x_0)|<\epsilon$$

holds.

Since $\lim \limits_{z \rightarrow x_0} f(z)= g(x_0)$ , then there exists a $\delta>0$ such that:

$$z\in \left ( x_0-\delta, x_0+\delta \right )\setminus \{x_0\}\Rightarrow \left | f(z)- g(x_0) \right |< \frac{\epsilon}{2}$$

or equivenantly:

$$z\in \left ( x_0-\delta, x_0+\delta \right )\setminus \{x_0\}\Rightarrow g(x_0)- \frac{\epsilon}{2}<f(z)< g(x_0)+ \frac{\epsilon}{2}$$

The above relation gives, as a consequence of the sandwich theorem, the following:

$$z\in \left ( x_0-\delta, x_0+\delta \right )\setminus \{x_0\}\Rightarrow g(x_0)- \frac{\epsilon}{2}<\lim_{y \rightarrow z}f(y)< g(x_0)+ \frac{\epsilon}{2}$$

and of course the result follows since the last quantity is less than $\frac{\epsilon}{2}$ and obviously less than $\epsilon$.