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My proof: 6.5.10. We will prove by induction that for each $k \in \mathbb{N} \cup \{0\}$, there exists $(c_n) \to 0$ such that for each $n \in \mathbb{N}$, $c_n \neq 0$ and $g^{(k)}(c_n) = 0$.

Base case, $k = 0$: We are given $(x_n) \to 0$, and for each $n \in \mathbb{N}$, $x_n \neq 0$ and $g(x_n) = 0$.

Inductive step: Let $k \in \mathbb{N} \cup \{0\}$ be arbitrary and let $(y_n) \to 0$ be the sequence corresponding to $k$ as in the inductive hypothesis. Since $g^{(k)}$ is continuous on $(-R, R)$, $g^{(k)}(0) = \lim_{n \to \infty}g^{(k)}(y_n) = \lim_{n \to \infty}0 = 0$. We now construct a sequence $(c_n)$. For each $n \in \mathbb{N}$: By the mean value theorem, there exists $c \in (0, y_n)$ such that $g^{(k + 1)}(c) = \frac{g^{(k)}(y_n) - g^{(k)}(0)}{y_n - 0} = 0$. Let $c_n = c$.

For each $n \in \mathbb{N}$, $0 < c_n < y_n$. By the squeeze theorem, $(c_n) \to 0$. By construction, $c_n \neq 0$ and $g^{(k + 1)}(c_n) = 0$. This completes the proof by induction.

The previous theorem and the continuity of all derivatives of $g$ implies that for each $n \in \mathbb{N} \cup \{0\}, g^{(n)}(0) = 0$. This implies that $g^{(n)}(0) = n!b_n = 0$ for each $n \in \mathbb{N}$, which means $b_n = 0$ for each $n \in \mathbb{N}$. Thus for each $x \in (-R, R)$, $g(x) = 0$.

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    $\begingroup$ This is a very nice proof. $\endgroup$
    – ProfOak
    Mar 22, 2020 at 20:11
  • $\begingroup$ Under the given circumstances one can prove that $g^{(k)} (0)=0$ for all $k=0,1,2,\dots$ (via mean value theorem). This means that $b_n=0$ for all $n$. $\endgroup$
    – Paramanand Singh
    Mar 23, 2020 at 3:05

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