$f(x)=\sum_{n=0}^{\infty}a_n x^n$ and there exists a sequence $x_n\to 0$, such that $f(x_n)=0$, for all $n$. Then $f\equiv 0$. I found this question really difficult for me, I don't even know how to start with it? Could you help me? I will appreciate that.

Prove that if $f(x)=\sum_{n=0}^{\infty}a_n x^n$ (defined in $(-R,R)$) and there exists a sequence $(x_n)$, with $x_n \neq 0$ for all $n$, tending to $0$ such that $f(x_n)=0$ for all $n$, then $f(x)=0$ for all $x$.

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 A: First of all, since $f$ is continuous, $f(0) = 0$, so $a_0 = 0$.
Now, think about $f(x)/x = \sum_{n=0}^\infty a_{n+1}x^n$, which is again continuous. For any $x_n$, we have $f(x_n)/x_n = 0$, so $f(x)/x \rightarrow 0$, as $x \rightarrow 0$. Hence, $a_1 = 0$.
Continuing this way, we show that $a_n = 0$ for all $n$.
Note that we need to assume that all $x_n$ are distinct from $0$.
A: By continuity of $f$ we have that $\,f(0)=\lim_{n\to\infty}f(x_n)=0$. 
As $f\in C^\infty(-R,R)$, there exist $y_n\in(x_{n+1},x_n)$, such that 
$$
0=\frac{f(x_{n+1})-f({x_n})}{x_{n+1}-x_n}=f'(y_n),
$$ 
due to Mean Value Theorem.
Clearly $y_n\to 0$, and due to continuiuty of $f'$, we have $f'(0)=\lim_{n\to\infty}f'(y_n)=0$.
Similarly, we may found a sequence $\{z_n\}$, with $z_n\to 0$, and $f''(z_n)=0$,
and hence $f''(0)=0$. 
Recursively, we obtain that $\,f^{(n)}(0)=0$, for all $n$, which implies that $a_n=0$, for all $n$, and hence $f\equiv 0$.
A: If $f(x)$ is not identically $0$, some $a_n$ is nonzero.  Suppose $a_m$ is the first nonzero coefficient.  Then $g(x) = f(x)/x^m = a_m + a_{m+1} x + \ldots$ is a convergent power series, therefore continuous, and $g(0) = a_m \ne 0$.
By continuity, there is $\epsilon > 0$ such that $g(x) \ne 0$ for $|x| < \epsilon$, and thus $f(x) \ne 0$ for $0 < |x| < \epsilon$.
