VE -1 lim エ→1 エー1 = 1/2
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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#121 part b
![**Mathematics - Limits and Continuity**
---
**Problem 120: Evaluating Limits using Definitions**
- **Task**: Use the definition of a limit to verify that
\[
\lim_{{x \to a}} \frac{{x^2 - a^2}}{{x - a}} = 2a.
\]
---
**Problem 121: Verifying Limits**
(a)
\[
\lim_{{x \to 1}} \frac{{x^3 - 1}}{{x - 1}} = 3
\]
**Hint**:
\[
\left| \frac{{x^3 - 1}}{{x - 1}} - 3 \right| = \left| x^2 + x + 1 - 3 \right|
\]
\[
\leq \left| x^2 - 1 \right| + \left| x - 1 \right|
\]
\[
= \left| (x - 1)(x + 1) \right| + \left| x - 1 \right|
\]
\[
= \left| (x - 1) \right| \cdot \left( \left| x + 1 \right| + 1 \right)
\]
\[
\leq \left| x - 1 \right|^2 + 3 \left| x - 1 \right|.
\]
(b)
\[
\lim_{{x \to 1}} \frac{{\sqrt{x} - 1}}{{x - 1}} = \frac{1}{2}
\]
---
**Continuity: What It Isn’t and What It Is**
**Hint**:
\[
\left| \frac{{\sqrt{x} - 1}}{{x - 1}} - \frac{1}{2} \right|
\]
\[
= \left| \frac{1}{{\sqrt{x} + 1}} - \frac{1}{2} \right|
\]
\[
= \frac{2 - (\sqrt{x} + 1)}{2(\sqrt{x} + 1)}
\]
\[
= \frac{1 - x}{2(1 + \sqrt{x})^2}
\]
\[
\leq \frac{1}{2} \left| x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc074fc7e-4e48-46f1-bfc4-aee6ba4e6262%2F1dd6d8fc-0520-47b9-8496-36256d937ab4%2F800wtir_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Mathematics - Limits and Continuity**
---
**Problem 120: Evaluating Limits using Definitions**
- **Task**: Use the definition of a limit to verify that
\[
\lim_{{x \to a}} \frac{{x^2 - a^2}}{{x - a}} = 2a.
\]
---
**Problem 121: Verifying Limits**
(a)
\[
\lim_{{x \to 1}} \frac{{x^3 - 1}}{{x - 1}} = 3
\]
**Hint**:
\[
\left| \frac{{x^3 - 1}}{{x - 1}} - 3 \right| = \left| x^2 + x + 1 - 3 \right|
\]
\[
\leq \left| x^2 - 1 \right| + \left| x - 1 \right|
\]
\[
= \left| (x - 1)(x + 1) \right| + \left| x - 1 \right|
\]
\[
= \left| (x - 1) \right| \cdot \left( \left| x + 1 \right| + 1 \right)
\]
\[
\leq \left| x - 1 \right|^2 + 3 \left| x - 1 \right|.
\]
(b)
\[
\lim_{{x \to 1}} \frac{{\sqrt{x} - 1}}{{x - 1}} = \frac{1}{2}
\]
---
**Continuity: What It Isn’t and What It Is**
**Hint**:
\[
\left| \frac{{\sqrt{x} - 1}}{{x - 1}} - \frac{1}{2} \right|
\]
\[
= \left| \frac{1}{{\sqrt{x} + 1}} - \frac{1}{2} \right|
\]
\[
= \frac{2 - (\sqrt{x} + 1)}{2(\sqrt{x} + 1)}
\]
\[
= \frac{1 - x}{2(1 + \sqrt{x})^2}
\]
\[
\leq \frac{1}{2} \left| x
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