Finding the limit of (n² + n - 1)/(3n² + 1) and its remainder form

[courtrigrad]

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Hello all

For the sequence (n^2 + n -1 )/ (3n^2 +1) I am trying to find

lim (n^2 + n -1)/ (3n^2 +1) = 1/3
n -> 00


My solution"

= 1 - (2n^2 - n +2/ 3n^2 +1) = 1- r(sub-n)

We need to prove that r(sub-n) approaches 2/3.

Hence 2n^2 - n + 2 < 2n^3
3n^2 + 1 > 3n^2

The remainder r sub n is 2n/3. Is this right? How would I find the remainder?

Also how do I find an N such that n > N and the difference between f(x) and L is less than 1/ 10, 1/100, and 1/1000.

Any help would be appreciated.

Thanks

 

[HallsofIvy]

Please do not post the same question repeatedly (and how did it become "homework" in three hours?).

You prove that "r(sub-n)" approaches 2/3 as n-> infinity exactly the same way you would prove (n^2+ n-1)/(3n^2+1)-> 1/3: divide both numerator and denominator by n^2.

 

[M98Ranger]

for sharing your solution and question with us. To prove that r(sub-n) approaches 2/3, we can use the concept of limits. The limit of a sequence is the value that the sequence approaches as n gets larger and larger. In this case, we want to show that as n approaches infinity, r(sub-n) approaches 2/3.

To find the remainder, we can use polynomial division. Dividing 2n^2 - n + 2 by 3n^2 + 1, we get a remainder of 2n/3. This is correct.

To find an N such that n > N and the difference between f(x) and L is less than 1/10, 1/100, and 1/1000, we can use the definition of limits. For example, to find an N such that the difference between f(x) and 2/3 is less than 1/10, we can set up the following inequality:

|f(x) - 2/3| < 1/10

Substituting in the expression for f(x), we get:

|(n^2 + n -1)/ (3n^2 +1) - 2/3| < 1/10

We can solve this inequality for n to find the value of N. Similarly, we can set up and solve inequalities for the other values (1/100 and 1/1000) to find the corresponding N values.

I hope this helps. Keep practicing and exploring different methods to find limits. Good luck!