Answered: sin(nz) ha(x) =

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter2: Equations And Inequalities

Section2.1: Equations

Problem 49E

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Question

Let Show that hn → 0 uniformly on R but that the sequence of derivatives (hn) diverges for every x ∈ R.

Expert Solution

Step 1

Given:

$h_{n} (x) = \frac{\sin (n x)}{\sqrt{n}}$ .

We have to show that $h_{n} \to 0$ uniformly on R but that the sequence of derivatives $(h_{n}^{'})$ diverges for every $x \in R$ .

Step 2

Now,

$\begin{array}{rcl} \lim_{n \to \infty} |h_{n} (x)| & = & \lim_{n \to \infty} |\frac{\sin (n x)}{\sqrt{n}}| \\ \leq & \lim_{n \to \infty} \frac{1}{|\sqrt{n}|} [∵ |\sin t| \leq 1] \\ = & 0 \end{array}$

Let $ε > 0$ , then there exists an integer $N \in ℕ$ such that $N > \frac{1}{ε^{2}}$ .

Then for all $x \in R$ and for all $n \geq N$ , $|h_{n} (x)| < \frac{1}{\sqrt{N}} < ε$ .

This implies that $h_{n} (x) \to 0$ uniformly on R.

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