How do I find the remainder and an N value for a sequence approaching 2/3?

[courtrigrad]

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(sun-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

 

[matt grime]

divide every element in the expression (use brackets!) by n^2, and use the fact that addition and so on are continuous (lim (a_n+b_n) = lima_n + limb_n when all three limits exist, as does: lim(a_n/b_n) = lim(a_n)/lim(b_n) provided b_n doesn't tend to zero

 

[HallsofIvy]

I assume that you mean lim (n->00) (n^2 + n -1)/(3n^2 +1) .

If that is the case then, dividing both numerator and denominator by n^2,
you get (1+ 1/n- 1/n^2)/(3+ 1/n) which is NOT 1- r_n.

What happens to 1/n and 1/n^2 as n-> 00?

 

[Pyrrhus]

Or, you could put it to its function equivalent, and then apply L'Hospital.

 

[courtrigrad]

No i do not want to use L' Hopitals Rule. First off:

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

The question tells me to show that r(sub-n) approaches 2/3. Hence we need to find the remainder r(sub-n). I tried finding the remainder by doing this:

Find an expression which is greater than (2n^2 + n -1).
Find an expression which is less than (3n^2 +1)

I got 2n^3 / 3n^2 = 2n/3. Now that I have found the remainder, can anyone verify that this is correct? Also I really need help in finding n such that n > N and the difference between f(x) and L is less than 1/10, 1/1000, 1/10000. I know that after getting the remainder, the difference between f(x) and L cannot be more than 2n/3. But I don't know what n is to satisfy the inequality.

Any helps would be greatly appreciated.

Thanks!

 

[courtrigrad]

Any help would be greatly appreciated!

 

[HallsofIvy]

Since you have completely ignored what help you have been given, I can't think of anything more to say.

 

[courtrigrad]

Method of Limits NOT SIMPLY RULES!

I am sorry HallsofIvy. However I was just wondering if I wanted to find a certain N such that the difference between N and the limit is less than for example 1/10. How would I use the remainder as a tool? I know that the remainder let's say for arguments sake is 2/n, then the difference between the function and the limit cannot exceed 2/n. Hence the main problem is whether I have to guess and check to find the number N, or could I use the remainder as a tool??


Any help would greatly be appreciated.



THanks

 

[courtrigrad]

I guess that you have to guess and check?

 

[courtrigrad]

For example if we want to find an n such that f(x) - L is less than 1/10, than n = 1. How do i find this? Do i guess and check?

 

[Kelvie]

You may want to do a forum search for delta-epsilon proofs. The way to do it for a limit at infinity is as follows (correct me if I'm wrong).

For:
<br /> \lim_{x \to \infty} f(x) = L<br />
Assuming L is a finite number, you want to find a N such that
<br /> |x| &gt; N(\epsilon) \implies |f(x) - L| &lt; \epsilon<br />

Or am I wrong?

As to how to find it.. just take out a stack of paper buy a few pencils..

And to your original question.. I just solved the inequality for n and got stuck at: (I'm not very good at math)
<br /> \begin{align*}<br /> &amp;4n^2\epsilon - 3n + 3\epsilon -4 &lt; 0 \\<br /> &amp;n &gt; \frac{3+\sqrt{9+16\epsilon-12 \epsilon^2}}{8\epsilon} \\<br /> &amp; or\\<br /> &amp;n &gt; \frac{3-\sqrt{9+16\epsilon-12 \epsilon^2}}{8\epsilon}<br /> \end{align*}<br />
One of those SHOULD be your N...

It's been a long time since I've done the quadratic equation with an inequality. Perhaps I shouldn't have done that? Anyone more learned than myself care to step in? :P

 

[HallsofIvy]

I am sorry HallsofIvy. However I was just wondering if I wanted to find a certain N such that the difference between N and the limit is less than for example 1/10. How would I use the remainder as a tool? I know that the remainder let's say for arguments sake is 2/n, then the difference between the function and the limit cannot exceed 2/n. Hence the main problem is whether I have to guess and check to find the number N, or could I use the remainder as a tool??


Any help would greatly be appreciated.



THanks

Okay, although that's not quite the question you originally asked!

First, it should be clear that the the limit of your original rational function is 2/3- you can see that by, as was originally suggested, dividing both numerator and denominator by n². (n^2 + n -1) / (3n^2 +1)
equals (1+ 1/n- 1/n²)/(3+ 1/n²). Obviously as n goes to infinity, each of those fractions with n in the denominator goes to 0 and so the whole fraction goes to 1/3. That's not a "rule", that's a perfectly reasonable use of the concept of "limit".

If you must for some reason use the definition of limit then, instead of writing the fraction as "1- another fraction" you really should write it as
"1/3- another fraction".

"Long division" of n^2+ n-1 by 3n^2+ 1 gives 1/3 + (n- 2/3)/(3n^2+ 1).

You want (n- 2/3)/(3n^2+ 1)< &epsilon; Since 3n^2+ 1 is always positive, we can multiply both sides of the inequality by 3n^2+ 1 to get
n- 2/3< 3&epsilon;n^2+ &epsilon or 3&epsilon;n²- n+ &epsilon;+ 2/3> 0. For what n is that true. It would be nice if we could factor it but we can, at any rate, solve 3&epsilon;n²- n+ &epsilon;+ 2/3= 0 using the quadratic equation. n= (1+/- &radic;(1- 4&epsilon;(&epsilon+2/3))/2. Messy, but doable! In the example you give, where &epsilon;= 0.1 this becomes n= (1+/- &radic(1- (.4)(.766)))/2.
I get n= 0.916 and n= .08. Since x has to be a positive integer that just tells us that ANY n will work! For smaller values of &epsilon; we might find larger values of n but there will still be some value that will work.

 

[Kelvie]

Hmm.. I got a different quadratic equation(above, probably wrong?).

Time to check over my work !

By the way.. when you get values for n.. does it mean that n > that value?

Wait.. after a quick google, in your last example, wouldn't n be bounded between the two values from the q. equation?

 

[wisky40]

I'm going to try to help you a little bit more
|(n^2+n-1)/(3n^2+1)-(1/3)|< or = |(3(n^2+n-1)-(3n^2+1))/3(3n^2+1)|<
or =|(3n-4)/3(3n^2+1)|<|(3n-3)/3(3n^2+1)|< or =|(n-1)/(3n^2+1)|<
|n/3n^2|< or= |1/3n|=1/3n<error => try to make the error = (1/10^k) =>
1/3n<1/10^k => n>(10^k)/3.
I hope this help you contrigrad.
wisky40