Prove $f(x)\equiv 0$

Question

Assume $f$ is $C^\infty$ on $\mathbb R$, and satisfies

1. There exists $M>0$ such that for any non-negetive integer $n$ and any $x\in\mathbb R$, $\left|f^{(n)}(x)\right|\leqslant M$
2. $\displaystyle f\left(\dfrac{1}{m}\right)=0, m\in \mathbb Z^{+}$

Prove: $f(x)\equiv 0$ on $\mathbb R$.

I have no ideas about this problem. How can we make use of the second claim? Does it implies that we can firstly prove $f(x)=0$ at every rational numbers and then extend to the whole real numbers?