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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![**Problem 7: Limit Calculation**
**Objective:**
Show all steps to find the limit as \( k \) approaches infinity for the expression \( \frac{\sin(e^{-k})}{e^{-k}} \) or demonstrate that the limit diverges.
### Solution Steps:
1. **Expression Analysis:**
The given limit is:
\[
\lim_{{k \to \infty}} \frac{\sin(e^{-k})}{e^{-k}}
\]
2. **Simplifying \( e^{-k} \) as \( k \) Approaches Infinity:**
- As \( k \) approaches infinity, \( e^{-k} \) approaches 0 since the exponential function decays rapidly.
3. **Substitution:**
Let \( x = e^{-k} \). When \( k \to \infty \), \( x \to 0^+ \).
The limit then becomes:
\[
\lim_{{x \to 0^+}} \frac{\sin(x)}{x}
\]
4. **Using Standard Limit Result:**
It is a well-known result in calculus that:
\[
\lim_{{x \to 0}} \frac{\sin(x)}{x} = 1
\]
5. **Applying the Limit Result:**
Therefore,
\[
\lim_{{k \to \infty}} \frac{\sin(e^{-k})}{e^{-k}} = 1
\]
### Conclusion:
The limit \( \lim_{{k \to \infty}} \frac{\sin(e^{-k})}{e^{-k}} \) is equal to 1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F39f80a36-5866-477b-b28d-154d74fc6779%2F872d48b0-8988-4ee2-8ddd-3781ff14a9f1%2Fbxm2x0r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 7: Limit Calculation**
**Objective:**
Show all steps to find the limit as \( k \) approaches infinity for the expression \( \frac{\sin(e^{-k})}{e^{-k}} \) or demonstrate that the limit diverges.
### Solution Steps:
1. **Expression Analysis:**
The given limit is:
\[
\lim_{{k \to \infty}} \frac{\sin(e^{-k})}{e^{-k}}
\]
2. **Simplifying \( e^{-k} \) as \( k \) Approaches Infinity:**
- As \( k \) approaches infinity, \( e^{-k} \) approaches 0 since the exponential function decays rapidly.
3. **Substitution:**
Let \( x = e^{-k} \). When \( k \to \infty \), \( x \to 0^+ \).
The limit then becomes:
\[
\lim_{{x \to 0^+}} \frac{\sin(x)}{x}
\]
4. **Using Standard Limit Result:**
It is a well-known result in calculus that:
\[
\lim_{{x \to 0}} \frac{\sin(x)}{x} = 1
\]
5. **Applying the Limit Result:**
Therefore,
\[
\lim_{{k \to \infty}} \frac{\sin(e^{-k})}{e^{-k}} = 1
\]
### Conclusion:
The limit \( \lim_{{k \to \infty}} \frac{\sin(e^{-k})}{e^{-k}} \) is equal to 1.
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