Solutions

a)

Consider

$$f(x)=\sin\left(\frac{1}{x}\right).$$

Notice that

$$\lim_{x\to 0}\sin\left(\frac{1}{x}\right)$$

does not exist, but by the given definition, for every $\delta>0$, we can choose $\varepsilon=1$ and have

$$0<|x|<\delta \Rightarrow |f(x)|<\varepsilon=1.$$

b)

Consider

$$f(x)= \begin{cases} 1, & x\geq 0,\\ -1, & \text{otherwise.} \end{cases}$$

Notice that $\lim_{x\to 0}f(x)$ does not exist. Let $l=0$. For $\varepsilon\leq 1$, the implication is vacuously true. For $\varepsilon>1$, the implication holds for any $\delta>0$.