# Indeterminate Forms and l'Hopital's Rule

1. Nov 6, 2012

### Reefy

1. The problem statement, all variables and given/known data
Lim as x→∞ of ((2x+1)/(2x-1))^(sqrtx)

2. Relevant equations

3. The attempt at a solution

When I initially plugged in ∞ for my x, I get (∞/∞)^∞, correct?

If so, should I just let y=((2x+1)/(2x-1))^(sqrtx) and take the limit of both sides using ln?

That's what I attempted to do and I got lim x→∞ of (sqrtx)(ln((2x+1)/(2x-1))^(sqrtx)) = (inf)(ln(∞/∞)) which I can't make any sense of.

Last edited: Nov 6, 2012
2. Nov 6, 2012

### SammyS

Staff Emeritus
Hello Reefy. Welcome to PF !

I'm not sure what you mean by: "take the limit of both sides using ln?"

You should consider taking the natural log of both sides of
$\displaystyle y=\left(\frac{2x+1}{2x-1}\right)^\sqrt{x}\ .$​
Then see if you can get a suitable indeterminate form so that you can use L'Hôpital's rule .

3. Nov 6, 2012

### Reefy

Thanks, I've been lurking for awhile and decided to make an account.

What I meant is exactly what you said lol. I wrote:

lim as x→∞ of lny = lim as x→∞ of ln((2x+1)/(2x-1))^(√x)

When I brought the √x down I got, lim as x→∞ of (√x)(ln((2x+1)/(2x-1)) = (∞) times ln(∞/∞).

Should I be getting ln (∞/∞) or am I doing something wrong? If it's correct, I don't know what to do from there on

Edit: How do I write out the equations and functions like you did so it can be clearer?

4. Nov 6, 2012

### Staff: Mentor

No. Your answer can't be ∞/∞, as that is indeterminate. Now, since (2x + 1)/(2x -1) → 2 as x → ∞, what you really have is [1], which is another indeterminate form.
That should be lim(x→∞) [√x * ln((2x+1)/(2x-1))]. Keep in mind that the expression here is different from what you started with - it's the log of that expression. Any limit you get will be the log of what you should have for the original expression.

5. Nov 6, 2012

### SammyS

Staff Emeritus
Last question first:

Those mathematical expressions are written using something called LaTeX. Here are some links which may help you with LaTeX.

https://www.physicsforums.com/misc/howtolatex.pdf

To get to the online LaTeX editor, click the Big Ʃ icon above the "Advanced" message box -- the box in which you compose your posts.

Now for the main subject. (I see that Mark44 has answered your post while I was composing the above.) I'll just make one point.

Write $\displaystyle \ \ \ln\left(\frac{2x+1}{2x-1}\right)\ \ \ \text{ as }\ \ \ \ln(2x+1)-\ln(2x-1)\ .$

6. Nov 6, 2012

### Reefy

Oops, I meant to take out that √x from the exponent. Thanks for pointing that out. I also tried the other way you showed with 1^∞ but didn't know what to do.

Oh, I forgot that I can rewrite the natural log. Lemme see if I can get the write answer now.