Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that $\lim_{y \to x} f(y)$ exists for every $x \in \mathbb{R}$. Prove that $g: \mathbb{R} \to \mathbb{R}: x \mapsto \lim_{y \to x} f(y)$ is continuous. (source problem: Spivak chapter 6)
My attempt:
Fix $a \in \mathbb{R}$.
Let $\epsilon > 0$. Choose $\delta > 0$ such that $|f(x)- g(a)| < \epsilon/2$ for all $x \in \mathbb{R}$ with $0 < |x-a| < \delta$ (such $\delta$ exists because $g(a) = \lim_{y \to a} f(y)$)
Fix $x \in \mathbb{R}$ with $|x-a| < \delta/2$.
Choose $\delta' > 0$ such that $|f(y)-g(x)| < \epsilon/2$ for all $y \in \mathbb{R}$ with $0 < |y-x| < \delta'$ (such $\delta'$ exists because $\lim_{y \to x} f(y) = g(x)$)
Pick $z \in \mathbb{R}- \{a\}$ satisfying $0 < |z-x| < \min \{\delta/2, \delta'\}$.
Then, we have $0 <|z-a| \leq |z-x|+|x-a| < \delta$, hence $|f(z)-g(a)| < \epsilon/2$ and we also have $|g(x)-f(z)|< \epsilon/2$.
Hence, for $x \in \mathbb{R}$ with $|x-a| < \delta/2$, we have $|g(x)-g(a)| \leq |g(x)-f(z)| + |f(z) - g(a)| < \epsilon$
and we are done.
Is this correct?