Solutions

In the forward direction, suppose

$$\lim_{x\to 0^+}f\left(\frac{1}{x}\right)=l.$$

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

$$0<x<\delta \Rightarrow \left|f\left(\frac{1}{x}\right)-l\right|<\varepsilon.$$

Let $z=1/x$. This gives

$$0<\frac{1}{\delta}<x \Longleftrightarrow 0<z<\delta \Rightarrow |f(z)-l|=|f(x)-l|<\varepsilon.$$

In the reverse direction $(\Leftarrow)$, suppose

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

Pick any $\varepsilon>0$. By definition, this means that there exists a $\delta>0$ such that, for all $x$,

$$x>\delta>0 \Rightarrow |f(x)-l|<\varepsilon.$$

Let $z=1/x$. Then,

$$\frac{1}{\delta}>x>0 \Longleftrightarrow z>\delta>0 \Rightarrow |f(z)-l|<\varepsilon,$$

which implies that

$$\lim_{z\to 0^+}f(z)=l.$$