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High School Math Calculus Precalculus with Limits: A Graphing Approach

The first five terms and limit and explain the limits exist or not.

The first five terms and limit and explain the limits exist or not.

Precalculus with Limits: A Graphing Approach

Precalculus with Limits: A Graphing Approach

7th Edition

ISBN: 9781305071711

Author: Ron Larson

Publisher: Brooks/Cole

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Concept explainers

Sequence and Series

A sequence is defined as a collection of things. Series is defined to sum the things one by one in the sequence. It was invented by German mathematician Carl Friedrich Gauss.

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Question

Chapter 11.4, Problem 45E

To determine

To calculate: The first five terms and limit and explain the limits exist or not.

Expert Solution & Answer

Answer to Problem 45E

The values of terms are $43,2,167,229, and 2811$ and limit is 3.

Explanation of Solution

Given information: $an=6n−22n+1 ...(1)$

To find the required five terms, put $n=1,2,3,4, and 5$ in equation (1), we get

Calculation:

At $n=1$

$a1=6×1−22×1+1 =6−22+1 =43$

At $n=2$

$a2=6×2−22×2+1 =12−24+1 =105 =2$

At $n=3$

$a3=6×3−22×3+1 =18−26+1 =167$

At $n=4$

$a4=6×4−22×4+1 =24−28+1 =229$

At $n=5$

$a1=6×5−22×5+1 =30−210+1 =2811$

Hence, first five terms are $43,2,167,229, and 2811$

Now, calculate the limit of equation (1), we get

Let, $∴limx→∞an=limx→∞6n−22n+1$

Formula used: $limx→a[f(x)g(x)]=limx→af(x)limx→ag(x), [∵limx→ag(x)≠0]$

$=limx→∞6n−22n+1$

Divide by highest denominator power

$=limn→∞ (6−2n2+1n)$

$=limn→∞ (6−2n)limn→∞ (2+1n)$

Put, the value of limit

$=(6−2∞2+1∞)=62=3$

$⇒limx→∞− (6n−22n+1)+limx→∞+ (6n−22n+1)=3$

Therefore, LHL is equal to RHL.

Hence, the limit is exist.

Chapter 11.4, Problem 44E

Chapter 11.4, Problem 46E

Chapter 11 Solutions

Precalculus with Limits: A Graphing Approach

Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Prob. 6E Ch. 11.1 - Prob. 7E Ch. 11.1 - Prob. 8E Ch. 11.1 - Prob. 9E Ch. 11.1 - Prob. 10E

Ch. 11.1 - Prob. 11E Ch. 11.1 - Prob. 12E Ch. 11.1 - Prob. 13E Ch. 11.1 - Prob. 14E Ch. 11.1 - Prob. 15E Ch. 11.1 - Prob. 16E Ch. 11.1 - Prob. 17E Ch. 11.1 - Prob. 18E Ch. 11.1 - Prob. 19E Ch. 11.1 - Prob. 20E Ch. 11.1 - Prob. 21E Ch. 11.1 - Prob. 22E Ch. 11.1 - Prob. 23E Ch. 11.1 - Prob. 24E Ch. 11.1 - Prob. 25E Ch. 11.1 - Prob. 26E Ch. 11.1 - Prob. 27E Ch. 11.1 - Prob. 28E Ch. 11.1 - Prob. 29E Ch. 11.1 - Prob. 30E Ch. 11.1 - Prob. 31E Ch. 11.1 - Prob. 32E Ch. 11.1 - Prob. 33E Ch. 11.1 - Prob. 34E Ch. 11.1 - Prob. 35E Ch. 11.1 - Prob. 36E Ch. 11.1 - Prob. 37E Ch. 11.1 - Prob. 38E Ch. 11.1 - Prob. 39E Ch. 11.1 - Prob. 40E Ch. 11.1 - Prob. 41E Ch. 11.1 - Prob. 42E Ch. 11.1 - Prob. 43E Ch. 11.1 - Prob. 44E Ch. 11.1 - Prob. 45E Ch. 11.1 - Prob. 46E Ch. 11.1 - Prob. 47E Ch. 11.1 - Prob. 48E Ch. 11.1 - Prob. 49E Ch. 11.1 - Prob. 50E Ch. 11.1 - Prob. 51E Ch. 11.1 - Prob. 52E Ch. 11.1 - Prob. 53E Ch. 11.1 - Prob. 54E Ch. 11.1 - Prob. 55E Ch. 11.1 - Prob. 56E Ch. 11.1 - Prob. 57E Ch. 11.1 - Prob. 58E Ch. 11.1 - Prob. 59E Ch. 11.1 - Prob. 60E Ch. 11.1 - Prob. 61E Ch. 11.1 - Prob. 62E Ch. 11.1 - Prob. 63E Ch. 11.1 - Prob. 64E Ch. 11.1 - Prob. 65E Ch. 11.1 - Prob. 66E Ch. 11.1 - Prob. 67E Ch. 11.1 - Prob. 68E Ch. 11.1 - Prob. 69E Ch. 11.1 - Prob. 70E Ch. 11.1 - Prob. 71E Ch. 11.1 - Prob. 72E Ch. 11.1 - Prob. 73E Ch. 11.1 - Prob. 74E Ch. 11.1 - Prob. 75E Ch. 11.1 - Prob. 76E Ch. 11.1 - Prob. 77E Ch. 11.1 - Prob. 78E Ch. 11.1 - Prob. 79E Ch. 11.1 - Prob. 80E Ch. 11.1 - Prob. 81E Ch. 11.1 - Prob. 82E Ch. 11.1 - Prob. 83E Ch. 11.1 - Prob. 84E Ch. 11.1 - Prob. 85E Ch. 11.1 - Prob. 86E Ch. 11.2 - Prob. 1E Ch. 11.2 - Prob. 2E Ch. 11.2 - Prob. 3E Ch. 11.2 - Prob. 4E Ch. 11.2 - Prob. 5E Ch. 11.2 - Prob. 6E Ch. 11.2 - Prob. 7E Ch. 11.2 - Prob. 8E Ch. 11.2 - Prob. 9E Ch. 11.2 - Prob. 10E Ch. 11.2 - Prob. 11E Ch. 11.2 - Prob. 12E Ch. 11.2 - Prob. 13E Ch. 11.2 - Prob. 14E Ch. 11.2 - Prob. 15E Ch. 11.2 - Prob. 16E Ch. 11.2 - Prob. 17E Ch. 11.2 - Prob. 18E Ch. 11.2 - Prob. 19E Ch. 11.2 - Prob. 20E Ch. 11.2 - Prob. 21E Ch. 11.2 - 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