# Indeterminate Forms and l'Hopital's Rule

Reefy

## 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.

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## 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 .

Reefy
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?

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.

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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/showthread.php?t=617570&highlight=latex

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

https://www.physicsforums.com/showthread.php?t=514469

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

Reefy
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/showthread.php?t=617570&highlight=latex

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

https://www.physicsforums.com/showthread.php?t=514469

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

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