# Indeterminate Forms and l'Hopital's Rule

## Homework Statement

Lim as x→∞ of ((2x+1)/(2x-1))^(sqrtx)

## 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:

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

Lim as x→∞ of ((2x+1)/(2x-1))^(sqrtx)

## 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.
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 .

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 .

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?

Mark44
Mentor

## Homework Statement

Lim as x→∞ of ((2x+1)/(2x-1))^(sqrtx)

## The Attempt at a Solution

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

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.

SammyS
Staff Emeritus
Homework Helper
Gold Member
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?
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 $\displaystyle \ \ \ln\left(\frac{2x+1}{2x-1}\right)\ \ \ \text{ as }\ \ \ \ln(2x+1)-\ln(2x-1)\ .$

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.

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.

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

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