Limits and sequences

**damani999** · May 3rd 2008, 02:27 PM

hi im stuck on a question and dnt knw what the answer is about..
Xn=(n^4+1)/(n^4-4). for n>2.
it asks me to determine L=lim n->infinity Xn
for each positive real number e, determine an integer No such that |Xn-L|<e for all integers n>No

**mr fantastic** · May 3rd 2008, 02:39 PM

Originally Posted by damani999

hi im stuck on a question and dnt knw what the answer is about..
Xn=(n^4+1)/(n^4-4). for n>2.
it asks me to determine L=lim n->infinity Xn
for each positive real number e, determine an integer No such that |Xn-L|<e for all integers n>No

I'll give some help with the first bit (needed for the second bit) and let you have another try at the second bit:

$\lim_{n \rightarrow \infty} \frac{n^4 + 1}{n^4 - 4} = \lim_{n \rightarrow \infty} \frac{1 + \frac{1}{n^4}}{1 - \frac{4}{n^4}} = \frac{1 + 0}{1 - 0} = 1$ .

**damani999** · May 3rd 2008, 02:45 PM

is it because 1/n^4 converges to 0 that the L=1. what do u do with this limit?

**mr fantastic** · May 3rd 2008, 02:57 PM

Originally Posted by damani999

is it because 1/n^4 converges to 0 that the L=1. what do u do with this limit?

Yes. You use it as the value of L in the second part. I'm sure your class notes or textbook will have an example to follow ......?

**damani999** · May 3rd 2008, 03:22 PM

nope.. my lecture notes arent very informant.. sorry bout that.
the answers are quite confusing. because they get 1+5/(n^4+1). dnt knw where that 5 came from?

**Plato** · May 3rd 2008, 03:37 PM

Originally Posted by damani999

the answers are quite confusing. because they get 1+5/(n^4+1). dnt knw where that 5 came from?

To get the answer in the text use simple "long division".
Divide $n^4 -4$ into $n^4 +1$ .

**mr fantastic** · May 3rd 2008, 03:44 PM

Originally Posted by damani999

nope.. my lecture notes arent very informant.. sorry bout that.
the answers are quite confusing. because they get 1+5/(n^4+1). dnt knw where that 5 came from?

Another way of doing it is to first note that $\frac{n^4 + 1}{n^4 - 1} = \frac{(n^4 -1) + 5}{n^4 - 1} = 1 + \frac{5}{n^4 - 1}$ ......

You're advised to thoroughly review the algebra that is pre-requisite for these sorts of questions.

**mr fantastic** · May 3rd 2008, 03:51 PM

Originally Posted by damani999

nope.. my lecture notes arent very informant.. sorry bout that.
the answers are quite confusing. because they get 1+5/(n^4+1). dnt knw where that 5 came from?

I'm sorry but I find it hard to believe that you would be given a question like this and have no example to refer to, either from class notes or textbook.

$|X_n - L| = \left| 1 + \frac{5}{n^4+1} - 1 \right| = \left|\frac{5}{n^4+1}\right| < \epsilon$ when $n > \left( \frac{5}{\epsilon} - 1\right)^{1/4}$ .

So choose $N_0 = \left( \frac{5}{\epsilon} - 1\right)^{1/4}$ .

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