is continuous for each x E R. Hint: Consider cases x 0 and x = 0. . For x0: Use the definition of continouity. i) Take € > 0 and assume that y x < 8, then ii) Explain why iii) Use the formula |u sin() - zsin (-) = |usin () - sin (;) y y x y - zx|sin() iv) Explain why ƒ (x) = { vi) Show that vii) Show that viii) Take 6 = min {4, 4 x sin (¹) 0 |x - y sin sin (a) sin (3) = 2 sin |yx| |yx| |xy| y sin (₁) if x #0 if x=0 sin (-:-) -sin (-) |≤|-|-|-2| |yx| |xy| v) Notice that if 6 <2, then using Example from Lecture 14 (see page) 8 explain why |y| > 2 3 (-)|- - y sin • For x = 0: Need to show that lima sin() = = 0. < 0, then show that for x 0 and y - x < 8 (-:-) = 270 a + B 2 1226. |x/20. -8 + 8 < €.
is continuous for each x E R. Hint: Consider cases x 0 and x = 0. . For x0: Use the definition of continouity. i) Take € > 0 and assume that y x < 8, then ii) Explain why iii) Use the formula |u sin() - zsin (-) = |usin () - sin (;) y y x y - zx|sin() iv) Explain why ƒ (x) = { vi) Show that vii) Show that viii) Take 6 = min {4, 4 x sin (¹) 0 |x - y sin sin (a) sin (3) = 2 sin |yx| |yx| |xy| y sin (₁) if x #0 if x=0 sin (-:-) -sin (-) |≤|-|-|-2| |yx| |xy| v) Notice that if 6 <2, then using Example from Lecture 14 (see page) 8 explain why |y| > 2 3 (-)|- - y sin • For x = 0: Need to show that lima sin() = = 0. < 0, then show that for x 0 and y - x < 8 (-:-) = 270 a + B 2 1226. |x/20. -8 + 8 < €.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Please answer all subparts to show that it is continuous. Thanks
![**5a. Continuity of Function \( f: \mathbb{R} \to \mathbb{R} \)**
**Function Definition:**
\[
f(x) =
\begin{cases}
x \sin \left( \frac{1}{x} \right) & \text{if } x \neq 0 \\
0 & \text{if } x = 0
\end{cases}
\]
We aim to demonstrate that \( f \) is continuous for each \( x \in \mathbb{R} \).
**Hint:** Consider the cases \( x \neq 0 \) and \( x = 0 \).
**Case \( x \neq 0 \):**
- Use the definition of continuity.
1. Take \( \epsilon > 0 \) and assume that \( |y-x| < \delta \), then:
\[
\left| y \sin \left( \frac{1}{y} \right) - x \sin \left( \frac{1}{x} \right) \right| = \left| y \sin \left( \frac{1}{y} \right) - x \sin \left( \frac{1}{y} \right) + x \sin \left( \frac{1}{y} \right) - x \sin \left( \frac{1}{x} \right) \right|
\]
\[
\leq |y-x| \left| \sin \left( \frac{1}{y} \right) \right| + |x-y| \left| \sin \left( \frac{1}{x} \right) \sin \left( \frac{1}{y} \right) \right|
\]
2. Explain why:
\[
|x-y| \left| \sin \left( \frac{1}{y} \right) \right| < \delta
\]
3. Use the formula:
\[
\sin(\alpha) - \sin(\beta) = 2 \sin \left( \frac{\alpha - \beta}{2} \right) \cos \left( \frac{\alpha + \beta}{2} \right)
\]
4. Explain why:
\[
\left| \sin \left(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa8d46346-fe9f-45aa-a3a0-df12b7cae379%2Fd490bfae-e1f5-4391-a9e6-1451a97279d8%2F18s421b_processed.png&w=3840&q=75)
Transcribed Image Text:**5a. Continuity of Function \( f: \mathbb{R} \to \mathbb{R} \)**
**Function Definition:**
\[
f(x) =
\begin{cases}
x \sin \left( \frac{1}{x} \right) & \text{if } x \neq 0 \\
0 & \text{if } x = 0
\end{cases}
\]
We aim to demonstrate that \( f \) is continuous for each \( x \in \mathbb{R} \).
**Hint:** Consider the cases \( x \neq 0 \) and \( x = 0 \).
**Case \( x \neq 0 \):**
- Use the definition of continuity.
1. Take \( \epsilon > 0 \) and assume that \( |y-x| < \delta \), then:
\[
\left| y \sin \left( \frac{1}{y} \right) - x \sin \left( \frac{1}{x} \right) \right| = \left| y \sin \left( \frac{1}{y} \right) - x \sin \left( \frac{1}{y} \right) + x \sin \left( \frac{1}{y} \right) - x \sin \left( \frac{1}{x} \right) \right|
\]
\[
\leq |y-x| \left| \sin \left( \frac{1}{y} \right) \right| + |x-y| \left| \sin \left( \frac{1}{x} \right) \sin \left( \frac{1}{y} \right) \right|
\]
2. Explain why:
\[
|x-y| \left| \sin \left( \frac{1}{y} \right) \right| < \delta
\]
3. Use the formula:
\[
\sin(\alpha) - \sin(\beta) = 2 \sin \left( \frac{\alpha - \beta}{2} \right) \cos \left( \frac{\alpha + \beta}{2} \right)
\]
4. Explain why:
\[
\left| \sin \left(
Expert Solution
Step 1: Recall the definition of continouity and a theorem.
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