Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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do no.6 pls and show full work
![### Advanced Calculus Problems and Solutions
#### 1. Limit Calculation Using L'Hospital's Rule
**Problem:** Find the limit. Say why L'Hospital's rule applies to this case and then use that rule to find the limit:
\[ \lim_{x \to 0} \frac{\sin x - x}{(e^{2x}) - 1} \]
**Solution:**
- Since \(\sin x - x \rightarrow 0\) and \((e^{2x} - 1) \rightarrow 0\) as \( x \rightarrow 0 \), we have an indeterminate form \( \frac{0}{0} \).
- L'Hospital's Rule applies to \( \frac{0}{0} \) indeterminate forms.
- Differentiate the numerator and denominator:
\[ \frac{d}{dx} (\sin x - x) = \cos x - 1 \]
\[ \frac{d}{dx} (e^{2x} - 1) = 2e^{2x} \]
- So, the limit becomes:
\[ \lim_{x \to 0} \frac{\cos x - 1}{2e^{2x}} \]
- Evaluate the limit:
\[ \frac{\cos 0 - 1}{2e^{0}} = \frac{1-1}{2 \cdot 1} = 0 \]
- Hence the limit is 0.
#### 2. Limit Calculation
**Problem:** Calculate the limit:
\[ \lim_{x \to \infty} \frac{\ln x}{\sqrt{x}} \]
**Solution:**
- As \( x \rightarrow \infty \), both \( \ln x \rightarrow \infty \) and \( \sqrt{x} \rightarrow \infty \).
- Apply L'Hospital's Rule:
\[ \frac{d}{dx} (\ln x) = \frac{1}{x} \]
\[ \frac{d}{dx} (\sqrt{x}) = \frac{1}{2\sqrt{x}} \]
- The limit becomes:
\[ \lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{1}{2\sqrt{x}}} = \lim_{x \to \infty} \frac{2\sqrt{x}}{x} = \lim_{x \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F92b3472e-ab59-40f2-b296-ac3727a4d9f3%2F116cc713-d498-4b3d-b0e1-5457ffe0fb00%2Fcvulk7f_processed.png&w=3840&q=75)
Transcribed Image Text:### Advanced Calculus Problems and Solutions
#### 1. Limit Calculation Using L'Hospital's Rule
**Problem:** Find the limit. Say why L'Hospital's rule applies to this case and then use that rule to find the limit:
\[ \lim_{x \to 0} \frac{\sin x - x}{(e^{2x}) - 1} \]
**Solution:**
- Since \(\sin x - x \rightarrow 0\) and \((e^{2x} - 1) \rightarrow 0\) as \( x \rightarrow 0 \), we have an indeterminate form \( \frac{0}{0} \).
- L'Hospital's Rule applies to \( \frac{0}{0} \) indeterminate forms.
- Differentiate the numerator and denominator:
\[ \frac{d}{dx} (\sin x - x) = \cos x - 1 \]
\[ \frac{d}{dx} (e^{2x} - 1) = 2e^{2x} \]
- So, the limit becomes:
\[ \lim_{x \to 0} \frac{\cos x - 1}{2e^{2x}} \]
- Evaluate the limit:
\[ \frac{\cos 0 - 1}{2e^{0}} = \frac{1-1}{2 \cdot 1} = 0 \]
- Hence the limit is 0.
#### 2. Limit Calculation
**Problem:** Calculate the limit:
\[ \lim_{x \to \infty} \frac{\ln x}{\sqrt{x}} \]
**Solution:**
- As \( x \rightarrow \infty \), both \( \ln x \rightarrow \infty \) and \( \sqrt{x} \rightarrow \infty \).
- Apply L'Hospital's Rule:
\[ \frac{d}{dx} (\ln x) = \frac{1}{x} \]
\[ \frac{d}{dx} (\sqrt{x}) = \frac{1}{2\sqrt{x}} \]
- The limit becomes:
\[ \lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{1}{2\sqrt{x}}} = \lim_{x \to \infty} \frac{2\sqrt{x}}{x} = \lim_{x \
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