1
$\begingroup$

Prove the sequence of functions $f_n(x) = \frac{\sin(nx)}{nx}$ converges pointwise on $(0, 1)$ to $f(x) = 0$.

I am trying to apply the the Squeeze theorem. Let $x \in (0 , 1)$. We know $-1 \leq \sin(x) \leq 1$ for all $x \in \mathbb{R}$ so we have $\frac{-1}{nx} \leq \frac{\sin(nx)}{nx} \leq \frac{1}{nx}$. I am trying to find another estimate to get rid of the $x$ value. Since $x \in (0,1)$, $\frac{1}{nx} \nleq \frac{1}{n}$ so I don't know where to go from here. Thanks.

  • 2
    Well, in any case you have $\lim_{n \to \infty} \frac{1}{nx}=0 \cdot \frac{1}{x} = 0$. – 2017-01-07
  • 1
    $\frac{-1}{nx} \leq \frac{\sin(nx)}{nx} \leq \frac{1}{nx} \hspace{1cm} \forall x \in (0,1)$. then take the limit as $n \rightarrow \infty$ – 2017-01-07

1 Answers 1

-1

In your estimate, $x$ should be fixed, then the both terms in the both sides converge to zero.