For each n ∈ N and x ∈ [0,∞), let (2)61+ x"and hn(x) = { ;{! if x2 1/nпх if0

Given: gnx=x1+xn and hnx=1if x≥1nnxif 0≤x<1n. To Do: (a) Find the pointwise limit on [0, ∞). (b)…

Answered: (2)6 1+ x" and hn(x) = { ; {! if x2 1/n…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

For each n ∈ N and x ∈ [0,∞), let

Expert Solution

Step 1

Given:

$g_{n} (x) = \frac{x}{1 + x^{n}}$ and $h_{n} (x) = \{\begin{cases} 1 & i f x \geq \frac{1}{n} \\ n x & i f 0 \leq x < \frac{1}{n} \end{cases}$ .

To Do:

(a) Find the pointwise limit on [0, $\infty$ ).

(b) Explain how the convergence is not uniform on [0, $\infty$ ).

Step 2

(a) $g_{n} (x) = \frac{x}{1 + x^{n}}$

For $0 \leq x < 1$ :

$\lim_{n \to \infty} x^{n} = 0$

And,
$\lim_{n \to \infty} g_{n} (x) = \lim_{n \to \infty} \frac{x}{1 + x^{n}} = x$ .

For $x = 1$ :

$\lim_{n \to \infty} 1^{n} = 1$

And,
$\lim_{n \to \infty} g_{n} (x) = \lim_{n \to \infty} \frac{1}{1 + 1^{n}} = \frac{1}{2}$ .

For $x > 1$ :

$\lim_{n \to \infty} x^{n} = \infty$

And,
$\lim_{n \to \infty} g_{n} (x) = \lim_{n \to \infty} \frac{x}{1 + x^{n}} = 0$ .

So, if $g (x)$ is the pointwise limit of $g_{n} (x)$ then, $g (x) = \{\begin{cases} x & i f 0 \leq x < 1 \\ \frac{1}{2} & i f x = 1 \\ 0 & i f x > 1 \end{cases}$ .

Now,
$h_{n} (x) = \{\begin{cases} 1 & i f x \geq \frac{1}{n} \\ n x & i f 0 \leq x < \frac{1}{n} \end{cases}$

For $x = 0$ :

$\lim_{n \to \infty} h_{n} (x) = \lim_{n \to \infty} 0 = 0$ .

For $x > 1$ :

As $\frac{1}{n} \to 0$ when $n \to \infty$ , so for $x > 1$ , there exists $N \in ℕ$ such that for all $n \geq N$ , $\frac{1}{n} \leq x$ .

Thus,
$\lim_{n \to \infty} h_{n} (x) = \lim_{n \to \infty n \geq N} h_{n} (x) = \lim_{n \to \infty n \geq N} 1 = 1$ .

If $h (x)$ is the pointwise limit of $h_{n} (x)$ then, $h (x) = \{\begin{cases} 0, & i f x = 0 \\ 1, & i f x > 0 \end{cases}$ .

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