Solutions

a)

Suppose

$$\lim_{x \to a} f(x)=l.$$

By definition,

$$\forall \epsilon > 0,\ \exists \delta > 0,\ 0<|x-a|<\delta \implies |f(x)-l|<\epsilon.$$

By the reverse triangle inequality, we have

$$||f(x)|-|l|| \le |f(x)-l| < \epsilon.$$

Hence,

$$\lim_{x \to a}|f(x)|=|l|.$$

b)

Suppose $\lim_{x \to a} f(x)=l$ and $\lim_{x \to a} g(x)=m$.
Without loss of generality, suppose also that $\max(l,m)=l$. For contradiction, assume that

$$\lim_{x \to a}\max(f,g)(x)=n \ne l.$$

Let $\epsilon=\left|\frac{l-n}{2}\right|$. By definition,

$$\begin{align*} \exists \delta_1,\quad 0<|x-a|<\delta_1 &\implies |f(x)-l|<\epsilon, \\ \exists \delta_2,\quad 0<|x-a|<\delta_2 &\implies |\max(f,g)(x)-n|<\epsilon. \end{align*}$$

Now, let $\delta=\min(\delta_1,\delta_2)$ and consider $x$ such that $0<|x-a|<\delta$. We have

$$-\left|\frac{l-n}{2}\right|+l < f(x) < \left|\frac{l-n}{2}\right|+l,$$$$-\left|\frac{l-n}{2}\right|+n < \max(f,g)(x) < \left|\frac{l-n}{2}\right|+n.$$

Consider two subcases: $l>n$ and $l<n$. If $l>n$, then $f(x)>\max(f,g)(x)$. If $l<n$, then either $\lim_{x \to a}g(x)=n$ or $f$ is not a function.

Hence,

$$\lim_{x \to a}\max(f,g)(x) = \max\left(\lim_{x \to a} f(x),\lim_{x \to a} g(x)\right).$$