because 0< < 1, the sequence converges monotonically to zero. 10.2.28 -0.003 < 1, so the sequence converges to zero. 10.2.29 Because |0.5| <1 and |0.75 < 1, the sequence converges to 0+3.0=0. en + " (:)") 10.2.30 lim lim (1+ which doesn't exist, because =1+ lim >1. > 1. 1. en n-00
because 0< < 1, the sequence converges monotonically to zero. 10.2.28 -0.003 < 1, so the sequence converges to zero. 10.2.29 Because |0.5| <1 and |0.75 < 1, the sequence converges to 0+3.0=0. en + " (:)") 10.2.30 lim lim (1+ which doesn't exist, because =1+ lim >1. > 1. 1. en n-00
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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What test are we using for # 30? The original problem is written on scrap paper in pencil
![### Sequence Convergence Analysis
#### Example 10.2.28
The sequence converges because \(0.003 < 1\), so the sequence converges to zero.
#### Example 10.2.29
Given that both \(0.5 < 1\) and \(0.75 < 1\), the sequence converges to \(0 + 3 \cdot 0 = 0\).
#### Example 10.2.30
To evaluate the limit:
\[
\lim_{n \to \infty} \frac{e^n + \pi^n}{e^n} = \lim_{n \to \infty} \left(1 + \left( \frac{\pi}{e} \right)^n \right) = 1 + \lim_{n \to \infty} \left( \frac{\pi}{e} \right)^n
\]
This limit doesn't exist because \(\left| \frac{\pi}{e} \right| > 1\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61b75115-d70f-4fe2-af93-2076876ad69a%2F3ba39d5a-5983-435a-8071-19ed6a3dbc6e%2Fljr109s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Sequence Convergence Analysis
#### Example 10.2.28
The sequence converges because \(0.003 < 1\), so the sequence converges to zero.
#### Example 10.2.29
Given that both \(0.5 < 1\) and \(0.75 < 1\), the sequence converges to \(0 + 3 \cdot 0 = 0\).
#### Example 10.2.30
To evaluate the limit:
\[
\lim_{n \to \infty} \frac{e^n + \pi^n}{e^n} = \lim_{n \to \infty} \left(1 + \left( \frac{\pi}{e} \right)^n \right) = 1 + \lim_{n \to \infty} \left( \frac{\pi}{e} \right)^n
\]
This limit doesn't exist because \(\left| \frac{\pi}{e} \right| > 1\).
![The image contains a mathematical expression written as a fraction.
**Expression:**
The numerator is \( e^{n + n} \).
The denominator is \( e^n \).
**Overall Expression:**
\[
\frac{e^{n+n}}{e^n}
\]
**Explanation:**
This expression simplifies the exponent rules. According to the laws of exponents, when you divide similar bases, you subtract the exponents. Therefore, this expression can be simplified as follows:
\[
\frac{e^{n+n}}{e^n} = e^{n+n-n} = e^n
\]
This results in \( e^n \) due to the simplification of the exponents.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61b75115-d70f-4fe2-af93-2076876ad69a%2F3ba39d5a-5983-435a-8071-19ed6a3dbc6e%2Fbk5byu6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The image contains a mathematical expression written as a fraction.
**Expression:**
The numerator is \( e^{n + n} \).
The denominator is \( e^n \).
**Overall Expression:**
\[
\frac{e^{n+n}}{e^n}
\]
**Explanation:**
This expression simplifies the exponent rules. According to the laws of exponents, when you divide similar bases, you subtract the exponents. Therefore, this expression can be simplified as follows:
\[
\frac{e^{n+n}}{e^n} = e^{n+n-n} = e^n
\]
This results in \( e^n \) due to the simplification of the exponents.
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