Exercise 1

Prove that the sequence { a }_{ n } = \frac { 2n + 4 }{ n } has a limit of 2. Also, calculate the terms whose distance from 2 is less than 0.1.

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Exercise 2

Prove that the sequence { a }_{ n } = \frac { 4n + 1 }{ n } has a limit of 4 and calculate how many terms of the succession are not within (4 - 0.001, 4 + 0.001).

Exercise 3

Prove that the sequence { a }_{ n } = \frac { { n }^{ 2 } }{ { n}^{ 2 } + 3 } has a limit of 1 and calculate how many terms of the succession are not within (1 - 0.001, 1 + 0.001).

Exercise 4

Prove that \lim_{ \frac { 3n - 8 }{ 4n + 1 } } = \frac { 3 }{ 4 }. Also, calculate the terms whose distance from the limit is less than 0.01.

Exercise 5

Prove that the sequence { a }_{ n } = \frac { { n }^{ 2 } + 1 }{ 4 } has a limit of + \infty and determine how many terms in the sequence are less than a million?

Exercise 6

Prove that the sequence { a }_{ n } = - { n }^{ 2 }a has a limit of - \infty. Also, what term of the sequence produces values of less than -10,000?

 

Solution of exercise 1

Prove that the sequence { a }_{ n } = \frac { 2n + 4 }{ n } has a limit of 2. Also, calculate the terms whose distance from 2 is less than 0.1.

\left | \frac { 2n + 4 }{ n } - 2 \right | < \frac { 1 }{ 10 }

\left | \frac { 2n + 4 - 2n }{ n } \right | < \frac { 1 }{ 10 }

\left | \frac { 4 }{ n } \right | < \frac { 1 }{ 10 }

\frac { 4 }{ n } < \frac { 1 }{ 10 }

n > 40

From { a }_{ 41 } the distance to 2 is less than 0.1

 

 

Solution of exercise 2

Prove that the sequence { a }_{ n } = \frac { 4n + 1 }{ n } has a limit of 4 and calculate how many terms of the succession are not within (4 - 0.001, 4 + 0.001).

\left | \frac { 4n + 1 }{ n } - 4 \right | < \frac { 1 }{ 1000 }

\left | \frac { 4n + 1 -4n }{ n } \right | < \frac { 1 }{ 1000 }

\left | \frac { 1 }{ n } \right | < \frac { 1 }{ 1000 }

\frac { 1 }{ n } < \frac { 1 }{ 1000 }

n > 1000

The first thousand terms of the sequence are out.

 

Solution of exercise 3

Prove that the sequence { a }_{ n } = \frac { { n }^{ 2 } }{ { n}^{ 2 } + 3 } has a limit of 1 and calculate how many terms of the succession are not within (1 - 0.001, 1 + 0.001).

\left | \frac { { n }^{ 2 } }{ { n }^{ 2 } + 3 } - 1 \right | < \frac { 1 }{ 1000 }

\left | \frac { -3 }{ { n }^{ 2 } + 3 } \right | < \frac { 1 }{ 1000 }

\frac { 3 }{ { n }^{ 2 } + 3 } < \frac { 1 }{ 1000 }

{ n }^{ 2 } + 3 > 3000

{ n }^{ 2 } > 2997

n > 54

The first 54 terms are out.

 

Solution of exercise 4

Prove that \lim_{ \frac { 3n - 8 }{ 4n + 1 } } = \frac { 3 }{ 4 }. Also, calculate the terms whose distance from the limit is less than 0.01.

\left | \frac { 3n - 8 }{ 4n + 1 } - \frac { 3 }{ 4 } \right | < \frac { 1 }{ 100 }

\left | \frac { 12n - 32 - 12n - 3 }{ 16n + 4 } \right | < \frac { 1 }{ 100 }

\left | \frac { -35 }{ 16n + 4 } \right | < \frac { 1 }{ 100 }

\frac { 35 }{ 16n + 4 }| < \frac { 1 }{ 100 }

16n + 4 > 3500

n > 218.5

From { n }_{ 219 } the distance to the limit is less than 0.01.

 

Solution of exercise 5

Prove that the sequence { a }_{ n } = \frac { { n }^{ 2 } + 1 }{ 4 } has a limit of + \infty and determine how many terms in the sequence are less than a million?

\frac { { n }^{ 2 } +  1 }{ 4 } > 1,000,000

n > \sqrt { 3,999,999 } \qquad n > 1,999

The 1,999 first terms of the sequence.

 

Solution of exercise 6

Prove that the sequence { a }_{ n } = - { n }^{ 2 }a has a limit of - \infty. Also, what term of the sequence produces values of less than -10,000?

-1, -4, -9, -16, -25, -36, -49, ...

- { n }^{ 2 } < - N \qquad n > \sqrt { N }

If N = 10,000 , its square root is 100, therefore, { a }_{ 101 }  a101 will be less than -10,000.

{ a }_{ 101 } = - { 101 }^{ 2 } = - 10,201

 

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Hamza

Hi! I am Hamza and I am from Pakistan. My hobbies are reading, writing and playing chess. Currently, I am a student enrolled in the Chemical Engineering Bachelor program.