Solutions

a) Suppose that $f$ is a function satisfying $|f(x)|\leq |x|$ for all $x$. Note that $f(0)=0$. Pick any $\varepsilon>0$. Let $\delta=\varepsilon-0.01$. For all $x$,

$$|x|<\delta \implies |f(x)|\leq |x|<\delta=\varepsilon-0.01<\varepsilon.$$

b) Consider

$$f(x)= \begin{cases} 0, & \text{if } x \text{ is rational},\\ x, & \text{if } x \text{ is irrational}. \end{cases}$$

c) Suppose that $g$ is continuous at $0$ and $g(0)=0$, and $|f(x)|\leq |g(x)|$.
Pick any $\varepsilon>0$. By definition, there exists $\delta>0$ such that

$$|x|<\delta \implies |f(x)|\leq |g(x)|<\varepsilon.$$

Hence

$$\lim_{x \to 0} f(x)=f(0)=0.$$