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High School Math Calculus Precalculus with Limits: A Graphing Approach

The five terms and limit of the sequence, if limit does not exist then give reason.

The five terms and limit of the sequence, if limit does not exist then give reason.

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

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topic…

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Question

Chapter 11.4, Problem 51E

To determine

To calculate:

The five terms and limit of the sequence, if limit does not exist then give reason.

Expert Solution & Answer

Answer to Problem 51E

First five terms are $1,−14,19,−116,125$ and limit of the sequence is $0$ .

Explanation of Solution

Given information:

$an=(−1)n+1n2$

Calculation:

For completing the table of the sequence,

Consider the given functions,

$an=(−1)n+1n2$

First five terms are : $a1,a2,a3,a4,a5$

For $n=1$ :

$an=(−1)1+112 =1$

For $n=2$ :

$an=(−1)2+122 =−14$

For $n=3$ :

$an=(−1)3+132 =19$

For $n=4$ :

$an=(−1)4+142 =−116$

For $n=5$ :

$an=(−1)5+152 =125$

Now for finding out the limit of the sequence in case it exists:

$limn→∞an=limn→∞(−1)n+1n2$

Use the property of limit,

$limn→∞f(x).g(x)=limn→∞f(x).limn→∞g(x)$

Thus,

$limn→∞(−1)n+1n2=limn→∞(−1)n+1.limn→∞1n2$

The limit $limn→∞1n2$ is $0$ as the degree of the numerator is less than the degree of the denominator.

Thus, the limit of the sequence is,

$limn→∞an=limn→∞(−1)n+1.limn→∞1n2 =limn→∞(−1)n+1.0 =0$

Chapter 11.4, Problem 50E

Chapter 11.4, Problem 52E

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 - 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