Proof that $\mathrm{min}(f,g)$ is continuous on $x_0 \in \mathbb{R}$ when $f, g$ are continuous on $x_0$ I already proved it using the formula
$$\min(f,g) = \frac{1}{2}(f+g) - \frac{1}{2} |f-g|$$
Now, I am trying to do it by an $\epsilon$-$\delta$ approach.
I have split the problem into two cases
$$f(x_0) = g(x_0)$$
and
$$f(x_0) < g(x_0)$$
Where $f(x_0) > g(x_0)$ is proved the same way as $f(x_0) < g(x_0)$.
I am stuck on how to proceed.
Thank you
 A: Let $x_0\in\mathbb R$, let $\varepsilon>0$, and take $\delta>0$ such that$$\lvert x-x_0\rvert<\delta\implies\bigl\lvert f(x)-f(x_0)\bigr\rvert,\bigl\lvert g(x)-g(x_0)\bigr\rvert<\frac\delta2.$$Then, if $\lvert x-x_0\rvert<\delta$,\begin{multline}\left\lvert\max\bigl(f(x),g(x)\bigr)-\max\bigl(f(x_0),g(x_0)\bigr)\right\rvert=\\=\left\lvert\frac{f(x)+g(x)-\bigl\lvert f(x)-g(x)\bigr\rvert}2-\frac{f(x_0)+g(x_0)-\bigl\lvert f(x_0)-g(x_0)\bigr\rvert}2\right\rvert=\\=\frac{\bigl\lvert f(x)-f(x_0)+g(x)-g(x_0)-\bigl\lvert f(x)-g(x)\bigr\rvert+\bigl\lvert f(x_0)-g(x_0)\bigr\rvert\bigr\rvert}2\leqslant\\\leqslant\frac{\bigl\lvert f(x)-f(x_0)\bigr\rvert+\bigl\lvert g(x)-g(x_0)\bigr\rvert+\left\lvert\bigl\lvert f(x)-g(x)\bigr\rvert-\bigl\lvert f(x_0)-g(x_0)\bigr\rvert\right\rvert}2\leqslant\\\leqslant\bigl\lvert f(x)-f(x_0)\bigr\rvert+\bigl\lvert g(x)-g(x_0)\bigr\rvert<\\<\delta.\end{multline}
A: Hint: if $f(x_0)<g(x_0)$, there is a neighborhood $U$ of $x_0$ such that
$$
f(x)<g(x).\quad\text{for every $x\in U$}
$$
