Limit of a sequence

1. Jan 18, 2017

kwal0203

1. The problem statement, all variables and given/known data

Suppose $$a_{n}=\frac{n^2-2n+1}{2n^2+4n-1}$$

For each positive number $$\epsilon$$, find a number N such that:

$$\mid a_{n} - L\mid < \epsilon$$ whenever n > N.

2. Relevant equations

3. The attempt at a solution

$$\mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1} {2} \mid < \epsilon$$

$$\mid \frac{3-8n} {4n^2+8n-2} \mid < \epsilon$$

Now I have no idea how to isolate the n so that I can find the N value. Any help appreciated.

Thanks!

Last edited: Jan 18, 2017
2. Jan 18, 2017

PeroK

Try some estimating.

Can you fix your latex?

3. Jan 18, 2017

kwal0203

Fixed it up.

Estimating what exactly?

4. Jan 18, 2017

PeroK

Estimating is a technique, which no one seems to teach explicitly for evaluating limits. Let me give you an example and see if you can apply the idea to your case.

Show that $5n^2 + 3n + 133$ tends to $\infty$ an $n \rightarrow \infty$

Estimating technique:

For $n > 1$ we have $n^2 > n$ hence $5n^2 > n$. So:

$5n^2 + 3n + 133 > 4n + 133 > 4n > n$

Now, we can use a simple $N$ in the limit argument. Whereas, trying to find $N$ for the orginal quadratic would be messy.

5. Jan 18, 2017

PeroK

PS I think you've lost a factor of $2$ from the denominantor.

6. Jan 18, 2017

kwal0203

Yes I definitely have, thanks.

7. Jan 18, 2017

kwal0203

So it goes kinda likes this:

For n > 1 we have $$n^2>n$$ hence, $$2n^2>n$$ So:

$$2n^2+4n-1>5n-1>5n>n$$

And in a similar way:

For n > 1 we have $$3 - 8n > 3 - 8 = -5$$ So:

$$\mid \frac{-5}{n} \mid < \epsilon$$

and,

$$n > 5 / \epsilon$$

Is that what you mean?

8. Jan 18, 2017

PeroK

First, you need to take that $\epsilon$ out of your working. That seems to be a common mistake to introduce $\epsilon$ into the inequality too soon. Second, your estimates have gone a bit wonky!

Let me show you what I mean. You should leave out the $\epsilon$ for now and do:

$\mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1} {2} \mid \ = \ \mid \frac{3-8n} {4n^2+8n-2} \mid \ < \ \mid \frac{8n} {4n^2+8n-2} \mid \ \dots$

Note that the numerator is less than $8n$ and you need an over-estimate on the numerator and an underestimate on the denominator. But, your underestimate on the denominator can't be too crude, because $8n/n$ doesn't converge to 0. So, you need an estimate based on $n^2$ in this case.

It's only when you've finished this algebra that you bring in $\epsilon$. For example:

Let $\epsilon > 0$ and take $N > 1/ \epsilon$ (or whatever you think you need from your calculations). Then $n > N$ implies:

$\mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1}{2} \mid < \frac{1}{n} < \epsilon$

9. Jan 18, 2017

kwal0203

Ok I think I get it now. Thanks for your help I'll do some more work on it.

10. Jan 18, 2017

PeroK

The simplest way to explain estimating is perhaps:

$|\frac{a(n)}{b(n)}| < |\frac{c(n)}{d(n)}|$

Where $|a(n)| < |c(n)|$ and $|b(n)| > |d(n)|$ and $c(n), d(n)$ are some simple expressions in $n$ that reduce to something that can easily be shown to have a limit of $0$.

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