Solutions

a) Suppose we have functions $g$ and $G$ such that

$$\lim_{x \to a}g(x)=\lim_{x \to a}G(x)=z.$$

Suppose further that

$$\lim_{y \to z}f(y)=l$$

for some function $f$.
Pick any $\varepsilon>0$. By definition, there exists $\delta>0$ such that for all $y$,

$$0<|y-z|<\delta \implies |f(y)-l|<\varepsilon.$$

There also exist $\alpha_1$ and $\alpha_2$ such that for all $x$,

$$0<|x-a|<\alpha_1 \implies |g(x)-z|<\delta$$

and

$$0<|x-a|<\alpha_2 \implies |G(x)-z|<\delta.$$

By picking $\alpha=\min(\alpha_1,\alpha_2)$, we have

$$0<|x-a|<\alpha \implies |g(x)-z|<\delta \implies |f(g(x))-l|<\varepsilon$$

and

$$0<|x-a|<\alpha \implies |G(x)-z|<\delta \implies |f(G(x))-l|<\varepsilon.$$

Hence,

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

Now, suppose $f$ is continuous at $l$ and $\lim_{x\to a}g(x)=l$. We define $G$ as

$$G(x)=g(x) \text{ for } x\ne a \quad \text{and} \quad G(a)=l.$$

Combining the result we just proved and the theorem on the composition of continuous functions, we have

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

b) Consider $f(x)=0$ for $x\ne l$ and $f(l)=1$. We have

$$\lim_{x \to a}f(g(x))=0\ne 1=f(l)=f\left(\lim_{x \to a}g(x)\right).$$