Answered: Question 6. In Example 2.4.3, show that…

Trigonometry (MindTap Course List)

10th Edition

ISBN:9781337278461

Author:Ron Larson

Publisher:Ron Larson

Chapter6: Topics In Analytic Geometry

Section6.4: Hyperbolas

Problem 5ECP: Repeat Example 5 when microphone A receives the sound 4 seconds before microphone B.

See similar textbooks

Related questions

Question

. In Example 2.4.3, show that fn(x) converges to 0 at each x ∈ [0, 1].

Define fn by
fn(x) = n X(0, 1
n ](x) = n if 0 < x ≤ 1
n ,
0 otherwise.
In this case limn→∞ fn(x) = 0 for all x ∈ [0, 1]. So here is an example
where the pointwise limit of a sequence of Lebesgue integrable functions is Lebesgue integrable. The odd thing is that 1
0
fn = 1 for
every n, but 1
0
0 = 0. In other words, here is an example where
limn→∞ b
a
fn =
b
a

limn→∞ fn

Question 6. In Example 2.4.3, show that fn (2) converges to 0 at each r € [0, 1].

Example 2.4.3. Define f. by
n if 0<r<
fa(x) = n X(@2(x) = {
%3D
(0.
otherwise.
In this case lim fn(x) 0 for all r E (0,1]. So here is an example
where the pointwise limit of a sequence of Lebesgue integrable fune-
tions is Lebesgue integrable. The odd thing is that fn = 1 for
every n, but
0 = 0. In other words, here is an example where
lim
What we are seeking are conditions that allow us to interchange
two limit-type operations, namely, integration (Lebesgue integration
in this case) and the limit of a sequence of functions. Our goal is a
theorem that addresses this issue, the Lebesgue Dominated Conver-
gence Theorem. Before the statement and proof of this theorem, we
need two lemmas.

Expert Solution

Step 1

Let us consider the function :

$f_{n} (x) = \{\begin{cases} n i f 0 < x ⩽ \frac{1}{n} \\ 0 o t h e r w i s e \end{cases}$

We have to prove that $f_{n} (x)$ converges to 0 for each $x \in [0, 1]$

By the definition of $f_{n} (x)$ for $x = 0$ , $f_{n} (x) = 0 \forall n \in ℕ$

Therefore, $f_{n} (0) \to 0 a s n \to \infty$ . ----------> (1)

Here we will first fix one $x_{0} \in (0, 1]$ and will show that $f_{n} (x_{0}) \to 0 a s n \to \infty$ . Now through the arbitraryness of the $x_{0} i n (0, 1]$ we can conclude $f_{n} (x) \to 0 a s n \to \infty \forall x \in [0, 1]$ [ since (1) also holds ]