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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![**Find the Limit**
Evaluate the following limit:
\[
\lim_{{x \to 0}} \frac{{\sin(3x)}}{{2x}}
\]
Use the limit property \(\lim_{{x \to 0}} \frac{{\sin(kx)}}{{kx}} = 1\) to solve, where \(k\) is a constant. Here, \(k = 3\).
**Solution:**
By multiplying and dividing by 3, the expression becomes:
\[
\lim_{{x \to 0}} \frac{{3\sin(3x)}}{{6x}} = \frac{3}{2} \lim_{{x \to 0}} \frac{{\sin(3x)}}{{3x}}
\]
Since \(\lim_{{x \to 0}} \frac{{\sin(3x)}}{{3x}} = 1\), the solution is:
\[
\frac{3}{2} \cdot 1 = \frac{3}{2}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70fe4f99-9e69-4bef-afe6-80d54e8903ee%2Fe3fa4514-be3c-49de-ab48-8678bbf7a961%2Fwo0wti9_processed.png&w=3840&q=75)
Transcribed Image Text:**Find the Limit**
Evaluate the following limit:
\[
\lim_{{x \to 0}} \frac{{\sin(3x)}}{{2x}}
\]
Use the limit property \(\lim_{{x \to 0}} \frac{{\sin(kx)}}{{kx}} = 1\) to solve, where \(k\) is a constant. Here, \(k = 3\).
**Solution:**
By multiplying and dividing by 3, the expression becomes:
\[
\lim_{{x \to 0}} \frac{{3\sin(3x)}}{{6x}} = \frac{3}{2} \lim_{{x \to 0}} \frac{{\sin(3x)}}{{3x}}
\]
Since \(\lim_{{x \to 0}} \frac{{\sin(3x)}}{{3x}} = 1\), the solution is:
\[
\frac{3}{2} \cdot 1 = \frac{3}{2}
\]
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