Solutions

a) Suppose that

$$\lim_{x \to a^+}f(x)=f(a)>0.$$

Let $\varepsilon=f(a)>0$. By definition, there exists $\delta>0$ such that for all $x$,

$$\begin{align*} 0\leq x-a<\delta &\implies |f(x)-f(a)|<f(a)\\ &\implies -f(a)<f(x)-f(a)<f(a)\\ &\implies 0<f(x)<2f(a). \end{align*}$$

The proof for the case where $f(a)<0$ is similar except that one can pick $\varepsilon=-f(a)>0$.

b) Proof is similar to that in a) except that the hypothesis is $0\leq b-x<\delta$ for some $\delta$.