Solutions

a)

Let

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

By definition,

$$\forall \varepsilon>0,\ \exists \delta>0, \quad 0<|x|<\delta \Rightarrow |g(x)|<\varepsilon.$$

Notice that $|\sin(1/x)|\leq 1$ for all $x$. Hence,

$$0<|x|<\delta \Rightarrow |g(x)||\sin(1/x)|<\varepsilon\cdot 1=\varepsilon,$$

and

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

b)

Let

$$\lim_{x\to 0}g(x)=0$$

and $|h(x)|\leq M$ for all $x$. If $M=0$, then $h(x)=0$ for all $x$ and

$$\lim_{x\to 0} g(x)h(x)=0$$

is trivially true. Consider $M>0$. Choose any $\varepsilon>0$. By definition, there exists a $\delta$ such that

$$0<|x|<\delta \Rightarrow |g(x)|<\frac{\varepsilon}{M}.$$

Thus,

$$|g(x)||h(x)|<\frac{\varepsilon}{M}\cdot M=\varepsilon.$$

Hence,

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