Prove that if $f(x)= e^{-1/x^2}\sin{\frac{1}{x}}$ for $x\neq0$ and $f(0)=0$, then $f^{(k)}(0)=0$ for all $k$. This question is from Spivak's Calculus (3rd ed) 18-41:

Prove that if $f(x)= e^{-1/x^2}\sin{\frac{1}{x}}$ for $x\neq0$ and $f(0)=0$, then $f^{(k)}(0)=0$ for all $k$.

Solution is:

I need someone to explain how $|\sin(1/x)|\leq1$ and $|\cos(1/x)|\leq1$ is helpful in the proof.  I can prove the same result for $f(x)= e^{-1/x^2}$ using the fact $\lim_{x \to \infty}\frac{x^n}{e^x}=0$ but not sure how to deal with the $\sin$ and $\cos$ in this question.  

I do not know if this is correct, but here is one approach to prove it.
Given that 
$$f^{\left( k \right)}\left( x \right) = e^{-\frac{1}{x^2}}\left[ \sum\limits_{i = 1}^{3k} \dfrac{a_i}{x^i} \sin \dfrac{1}{x} + \sum\limits_{i = 1}^{3k} \dfrac{b_i}{x^i} \cos \dfrac{1}{x} \right]$$
We have from here,
$$f^{\left( k \right)}\left( x \right) = e^{-\frac{1}{x^2}} \left[ \dfrac{a_1 \sin \dfrac{1}{x} + b_1 \cos \dfrac{1}{x}}{x} + \dfrac{a_2 \sin \dfrac{1}{x} + b_2 \cos \dfrac{1}{x}}{x^2} + \\ \dfrac{a_3 \sin \dfrac{1}{x} + b_3 \cos \dfrac{1}{x}}{x^3} + \dots + \dfrac{a_{3k} \sin \dfrac{1}{x} + b_{3k} \cos \dfrac{1}{x}}{x^{3k}} \right]$$
$$\therefore f^{\left( k \right)}\left( x \right) = \dfrac{e^{-\frac{1}{x^2}}}{x^{3k}} \left[ x^{3k - 1} \left( a_1 \sin \dfrac{1}{x} + b_1  \cos \dfrac{1}{x}\right) + x^{3k - 2} \left( a_2 \sin \dfrac{1}{x} + b_2  \cos \dfrac{1}{x}\right) + \\ x^{3k - 3} \left( a_2 \sin \dfrac{1}{x} + b_3  \cos \dfrac{1}{x}\right) + \dots + \left( a_{3k} \sin \dfrac{1}{x} + b_{3k}  \cos \dfrac{1}{x}\right) \right]$$
From here, we may conclude that $\lim\limits_{x \rightarrow 0} f^{\left( k \right)} \left( x \right) = 0$ which holds true for all positive k
