7. Let $f:[0,1) \rightarrow \mathbb{R}$ be a function that satisfies the following condition: if
$$
x=\sum_{n=1}^{\infty} \frac{a_{n}}{10^{n}}=. a_{1} a_{2} a_{3} \ldots
$$
is the decimal expansion of $x$ and there does not exist a positive integer $k$ such that $a_{n}=9$ for all $n \geq k$, then
$$
f(x)=\sum_{n=1}^{\infty} \frac{a_{n}}{10^{2 n}} .
$$

Determine $f^{\prime}\left(\frac{1}{3}\right)$.
Answer: 0 Note that $\frac{1}{3}=\sum_{n=1}^{\infty} \frac{3}{10^{n}}$.
Clearly $f$ is an increasing function. Also for any integer $n \geq 1$, we see from decimal expansions that $f\left(\frac{1}{3} \pm \frac{1}{10^{n}}\right)-f\left(\frac{1}{3}\right)= \pm \frac{1}{10^{2 n}}$.
Consider $h$ such that $10^{-n-1} \leq|h|<10^{-n}$. The two properties of $f$ outlined above show that $\left\lvert\, f\left(\frac{1}{3}+\right.\right.$ $h)-f\left(\frac{1}{3}\right) \left\lvert\,<\frac{1}{10^{2 n}}\right.$. And from $\left|\frac{1}{h}\right| \leq 10^{n+1}$, we get $\left|\frac{f\left(\frac{1}{3}+h\right)-f\left(\frac{1}{3}\right)}{h}\right|<\frac{1}{10^{n-1}}$. Taking $n \rightarrow \infty$ gives $h \rightarrow 0$ and $f^{\prime}\left(\frac{1}{3}\right)=\lim _{n \rightarrow \infty} \frac{1}{10^{n-1}}=0$.