The ordinary $\varepsilon$-$\delta$ definiton gives us $\delta_n> 0$ such that $|x-x_0|<\delta_n$ implies $|f(x)-L|< 1/n$. Now define recursively $\rho_1 = \delta_1$ and $\rho_{n+1} = \min(\rho_n/2, \delta_{n+1})$. Then $(\rho_n)$ is a strictly decreasing sequence of positive numbers (note $\rho_{n+1} \le \rho_n/2$). We also have $\rho_{n+1} \le 2^{-n}\rho_1$, thus $\bigcup_{n=2}^\infty [\rho_{n+1},\rho_n) = (0, \rho_2)$. Let $\Delta = \rho_2, E = 1$. Note that for each $\delta \in (0,\Delta)$ there exists a unique $n \ge 2$ such that $\delta \in [\rho_{n+1},\rho_n)$. Now define
$$h :(0,\Delta) \to (0,E), h(t) = \frac{1}{n} + \frac{t- \rho_{n+1}}{(n-1)n(\rho_n - \rho_{n+1})} \text{ for } t \in [\rho_{n+1},\rho_n) .$$
Clearly $h(\rho_{n+1}) = \frac{1}{n}$ and $[\rho_{n+1},\rho_n)$ is mapped strictly increasing onto $[\frac{1}{n}, \frac{1}{n-1})$, thus $h$ maps $(0,\Delta)$ strictly increasing onto $(0,1)$.
Let $\delta \in (0,\Delta)$ and $|x-x_0|<\delta$. Take $n \ge 2$ such that $\delta \in [\rho_{n+1},\rho_n)$. This shows that $\delta < \rho_n \le \delta_n$. Hence $|x-x_0|<\delta_n$ which implies
$$|f(x)-L|< 1/n = h(\rho_{n+1}) \le h(\delta) .$$
