Hello Reefy. Welcome to PF !Reefy said:Homework Statement
Lim as x→∞ of ((2x+1)/(2x-1))^(sqrtx)
Homework Equations
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.
SammyS said: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[itex]\displaystyleThen see if you can get a suitable indeterminate form so that you can use L'Hôpital's rule .
y=\left(\frac{2x+1}{2x-1}\right)^\sqrt{x}\ .[/itex]
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.Reefy said:Homework Statement
Lim as x→∞ of ((2x+1)/(2x-1))^(sqrtx)
Homework Equations
The Attempt at a Solution
When I initially plugged in ∞ for my x, I get (∞/∞)^∞, correct?
Reefy said: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 question first:Reefy said: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 said: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.
SammyS said: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 [itex]\displaystyle \ \ \ln\left(\frac{2x+1}{2x-1}\right)\ \ \ \text{ as }\ \ \ \ln(2x+1)-\ln(2x-1)\ .[/itex]
Indeterminate forms are mathematical expressions that cannot be evaluated using basic algebraic techniques. They typically involve the division of two quantities, where both the numerator and denominator approach either zero or infinity.
l'Hopital's Rule is a mathematical tool used to evaluate indeterminate forms. It states that if the limit of a function f(x) divided by g(x) is an indeterminate form, then the limit of f(x) divided by g(x) is equal to the limit of the derivatives of f(x) and g(x) divided by their derivatives.
l'Hopital's Rule should only be used when dealing with indeterminate forms in limits. It cannot be used for other types of mathematical expressions.
Some common indeterminate forms are 0/0, ∞/∞, 0∞, ∞ - ∞, and 1∞. These forms can arise from various mathematical expressions and cannot be solved using basic algebraic techniques.
Yes, there are limitations to l'Hopital's Rule. It can only be used for limits involving indeterminate forms. It cannot be used for limits that do not have an indeterminate form or for limits that have multiple variables.