Solutions

a) It suffices to prove that $g(x)=|x|$ is continuous. It follows by the theorem on the composition of continuous functions that $|f|$ is continuous at $a$ if $f$ is so. Hence, consider any $a\in\mathbb{R}$. Pick any $\varepsilon>0$. Let $\delta=\varepsilon$. It follows that for all $x$,

$$|x-a|<\delta \implies \big||x|-|a|\big|\leq |x-a|<\varepsilon.$$

and so

$$\lim_{x \to a}|x|=|a|.$$

b) Let

$$E(x)=\frac{f(x)+f(-x)}{2}$$

and

$$O(x)=\frac{f(x)-f(-x)}{2}.$$

Notice the functions are even and odd, respectively, and that both are continuous. In addition, $f=E+O$.

c) We have the identities

$$\max(f,g)(x)=\frac{f(x)+g(x)}{2}+\frac{|f(x)-g(x)|}{2}$$

and

$$\min(f,g)(x)=\frac{f(x)+g(x)}{2}-\frac{|f(x)-g(x)|}{2}.$$

Combining the result in a) and the theorem on the sum of continuous functions, we have

$$\begin{align*} \lim_{x \to a}\max(f,g)(x) &=\lim_{x \to a}\left[\frac{f(x)+g(x)}{2}+\frac{|f(x)-g(x)|}{2}\right]\\ &=\frac{f(a)+g(a)}{2}+\frac{|f(a)-g(a)|}{2}, \end{align*}$$

and

$$\begin{align*} \lim_{x \to a}\min(f,g)(x) &=\lim_{x \to a}\left[\frac{f(x)+g(x)}{2}-\frac{|f(x)-g(x)|}{2}\right]\\ &=\frac{f(a)+g(a)}{2}-\frac{|f(a)-g(a)|}{2}. \end{align*}$$

d) Let $g(x)=\max(0,f)(x)$ and $h(x)=\max(0,-f(x))$. Both functions are continuous and nonnegative. Also, $f=g-h$.