# I Understanding limits

1. Feb 6, 2017

### Jarvis88

I'm trying to understand why the $\lim_{n \rightarrow \infty} ( \displaystyle \frac {n+1}{n-1} )$ equals the indeterminate form 1?

I ask because we have started going over sequences and it was used as an example. I understand how to go from here- taking the ln of both sides and using L'Hopital's rule to get the limit of e2.

2. Feb 6, 2017

### pwsnafu

Could you write down what you teacher said in full. I have no idea what is going on.

The standard way to evaluate the limit is just observe $\frac{n+1}{n-1} = \frac{n-1+2}{n-1} = 1 + \frac2{n-1}$ and then it becomes obvious.

Edit: alternative $\frac{n+1}{n-1} = \frac{1+n^{-1}}{1-n^{-1}}$ and then apply limit laws. Again the limit is 1.

3. Feb 6, 2017

### Jarvis88

We were discussing using L'Hopital's rule for indeterminate forms to evaluate the sequence below. I just don't understand how that limit is
1. I guess I'm still fuzzy on how to figure out limits?

4. Feb 6, 2017

### pwsnafu

Oh, the limit you want to evalute is $\left(\frac{n+1}{n-1}\right)^{n}$ and not $\frac{n+1}{n-1}$. Yes that's $1^\infty$ indeterminate form.
This is what happens when students do it naively:

$\lim_{n\to\infty}\left(\frac{n+1}{n-1}\right)^{n} = \left(\lim_{n\to\infty}\frac{n+1}{n-1}\right)^{\lim_{n\to\infty} n} = \left(1+\lim_{n\to\infty}\frac2{n-1}\right)^{\lim_{n\to\infty} n} = 1^\infty$.

5. Feb 6, 2017

### Jarvis88

Thank you so much for the explanation!!!!

6. Feb 7, 2017

### Staff: Mentor

It doesn't. The limit here, which apparently isn't the one you really meant to ask about, is 1.
$\lim_{n \to \infty}\frac{n + 1}{n - 1} = \lim_{n \to \infty}\frac n n \frac{1 + 1/n}{1 - 1/n} = 1$
For any finite value of n, n/n is 1, and as n gets larger, the other fraction approaches 1 in value, making the limit equal to 1.