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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![The mathematical expression provided demonstrates the calculation of a limit as \( x \) approaches infinity. Specifically, the expression is:
\[ \lim_{{x \to \infty}} \frac{3e^{4x} - 2e^{-4x}}{5e^{4x} + e^{-4x}} \]
Here is a breakdown of the components of this expression:
- The notation \[ \lim_{{x \to \infty}} \] denotes the limit of the function as \( x \) approaches infinity.
- The fraction: \(\frac{3e^{4x} - 2e^{-4x}}{5e^{4x} + e^{-4x}}\)
Let’s analyze this limit step by step.
1. Identify the exponential terms:
- \( e^{4x} \) grows very rapidly as \( x \) approaches infinity.
- \( e^{-4x} \) diminishes very rapidly as \( x \) approaches infinity.
2. When \( x \) approaches infinity:
- \( e^{4x} \) will dominate both the numerator and the denominator because \( e^{-4x} \) will approach zero.
3. Simplify the expression by dividing both the numerator and denominator by \( e^{4x} \):
\[ \frac{3e^{4x}/e^{4x} - 2e^{-4x}/e^{4x}}{5e^{4x}/e^{4x} + e^{-4x}/e^{4x}} = \frac{3 - 2e^{-8x}}{5 + e^{-8x}} \]
4. As \( x \) approaches infinity, \( e^{-8x} \) approaches zero:
\[ \frac{3 - 0}{5 + 0} = \frac{3}{5} \]
Therefore, the limit is:
\[ \boxed{\frac{3}{5}} \]
This demonstrates how we can handle limits involving exponential functions by recognizing the dominant terms as the variable approaches infinity.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffe7817cb-5274-4434-90ec-13ab480c973d%2F7ccc86a4-4c90-4459-80c4-2a1b148ddb24%2Fz811m2_processed.png&w=3840&q=75)
Transcribed Image Text:The mathematical expression provided demonstrates the calculation of a limit as \( x \) approaches infinity. Specifically, the expression is:
\[ \lim_{{x \to \infty}} \frac{3e^{4x} - 2e^{-4x}}{5e^{4x} + e^{-4x}} \]
Here is a breakdown of the components of this expression:
- The notation \[ \lim_{{x \to \infty}} \] denotes the limit of the function as \( x \) approaches infinity.
- The fraction: \(\frac{3e^{4x} - 2e^{-4x}}{5e^{4x} + e^{-4x}}\)
Let’s analyze this limit step by step.
1. Identify the exponential terms:
- \( e^{4x} \) grows very rapidly as \( x \) approaches infinity.
- \( e^{-4x} \) diminishes very rapidly as \( x \) approaches infinity.
2. When \( x \) approaches infinity:
- \( e^{4x} \) will dominate both the numerator and the denominator because \( e^{-4x} \) will approach zero.
3. Simplify the expression by dividing both the numerator and denominator by \( e^{4x} \):
\[ \frac{3e^{4x}/e^{4x} - 2e^{-4x}/e^{4x}}{5e^{4x}/e^{4x} + e^{-4x}/e^{4x}} = \frac{3 - 2e^{-8x}}{5 + e^{-8x}} \]
4. As \( x \) approaches infinity, \( e^{-8x} \) approaches zero:
\[ \frac{3 - 0}{5 + 0} = \frac{3}{5} \]
Therefore, the limit is:
\[ \boxed{\frac{3}{5}} \]
This demonstrates how we can handle limits involving exponential functions by recognizing the dominant terms as the variable approaches infinity.
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