TITLE: Proving that: $\forall n \in \mathbb{N} :|f^{(n)}(x)|\leq \frac{1}{n+1}$
QUESTION [10 upvotes]: Suppose $x>0$ and we have function $f(x)=\frac{\sin x}{x}$ how can we show $$\forall n \in \mathbb{N} :|f^{(n)}(x)|\leq \frac{1}{n+1}$$ I need a hint to show this property .Thanks in advance . I tried for $n=1 ,2$  by finding maximum of $|f'| ,|f''|$ but I get stuck to show for $n$

REPLY [2 votes]: See that,  $\displaystyle\frac{\sin x}{x} =f(x) = \frac{1}{2}\int_{-1}^{1} e^{-itx} dt$ Then 
$$|f^{(n)}(x)| =\left|\frac{1}{2}\int_{-1}^{1} (-it)^ne^{-itx} dt\right|  \le\frac{1}{2}\int_{-1}^{1} |t|^n dt=\int_0^1t^n\,dt=\frac1{n+1}.$$