Solutions

a)

We define

$$\lim_{x\to a^+}f(x)=\infty$$

to mean that for all $N>0$, there is a $\delta>0$ such that, for all $x$, if $0<x-a<\delta$, then $f(x)>N$.

We define

$$\lim_{x\to a^-}f(x)=\infty$$

to mean that for all $N>0$, there is a $\delta>0$ such that, for all $x$, if $0<a-x<\delta$, then $f(x)>N$.

b)

Pick $N>0$. Let

$$\delta=\frac{1}{N}.$$

We have, for all $x$,

$$0<x<\frac{1}{N} \Rightarrow N<\frac{1}{x}.$$

c)

In the forward direction $(\Rightarrow)$, suppose

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

Pick $N>0$. By definition, there exists $\delta>0$ such that, for all $x$,

$$0<x<\delta \Rightarrow f(x)>N.$$

Let $z=1/x$. Then,

$$0<\frac{1}{\delta}<x \Longleftrightarrow 0<z<\delta \Rightarrow f(z)=f\left(\frac{1}{x}\right)>N.$$

The proof in the reverse direction is similar.