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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Question
![### Evaluate the Limit
\[ \lim_{x \to 0} \frac{\sin 2x}{6x} \]
This is a limit problem that requires evaluating the expression as \( x \) approaches 0. You can use the standard limit law:
\[ \lim_{x \to 0} \frac{\sin kx}{kx} = 1 \]
In this case, let \( k = 2 \). To solve the given expression:
1. Multiply the numerator and the denominator by 2 to match the form of the standard limit law:
\[ \lim_{x \to 0} \frac{\sin 2x}{6x} = \lim_{x \to 0} \frac{2(\sin 2x)}{12x} = \frac{1}{6} \lim_{x \to 0} \frac{\sin 2x}{2x} \]
2. Apply the standard limit law:
\[ \lim_{x \to 0} \frac{\sin 2x}{2x} = 1 \]
Therefore, the result is:
\[ \frac{1}{6} \times 1 = \frac{1}{6} \]
The limit evaluates to \(\frac{1}{6}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F621ac9f8-2dbb-4715-ab7b-42c3ebc90b44%2F561f77db-5f21-4c19-b502-c0049b2aa059%2Feo4abo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Evaluate the Limit
\[ \lim_{x \to 0} \frac{\sin 2x}{6x} \]
This is a limit problem that requires evaluating the expression as \( x \) approaches 0. You can use the standard limit law:
\[ \lim_{x \to 0} \frac{\sin kx}{kx} = 1 \]
In this case, let \( k = 2 \). To solve the given expression:
1. Multiply the numerator and the denominator by 2 to match the form of the standard limit law:
\[ \lim_{x \to 0} \frac{\sin 2x}{6x} = \lim_{x \to 0} \frac{2(\sin 2x)}{12x} = \frac{1}{6} \lim_{x \to 0} \frac{\sin 2x}{2x} \]
2. Apply the standard limit law:
\[ \lim_{x \to 0} \frac{\sin 2x}{2x} = 1 \]
Therefore, the result is:
\[ \frac{1}{6} \times 1 = \frac{1}{6} \]
The limit evaluates to \(\frac{1}{6}\).
Expert Solution
Step 1: We have to find the limit of the function.
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