A seemingly tricky limit - L'Hopital

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Homework Help Overview

The discussion revolves around evaluating the limit as \( t \) approaches infinity for the expression \( \frac {1-\frac{t}{(t-1)}}{1-\sqrt{\frac{t}{(t-1)}} \). The subject area includes limits and the application of L'Hôpital's rule.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore different methods for evaluating the limit, including L'Hôpital's rule and substitution techniques. Some question the necessity of taking derivatives, while others suggest alternative approaches like using difference of squares.

Discussion Status

The discussion includes various perspectives on the methods used to solve the limit problem. Some participants express approval of the original poster's approach, while others suggest simpler methods. There is an acknowledgment of differing classroom rules regarding the use of L'Hôpital's rule.

Contextual Notes

One participant notes that their class does not allow the use of L'Hôpital's rule, which influences their approach to the problem. This highlights the constraints under which some participants are operating.

Asphyxiated
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Homework Statement



[tex]\lim_{t \to \infty} \frac {1-\frac{t}{(t-1)}}{1-\sqrt{\frac{t}{(t-1)}}}[/tex]

Homework Equations





The Attempt at a Solution


I am pretty sure that everything I did here was legal, I just wanted to check. I got the right answer, so yeah, here's what I did:

[tex]\lim_{t \to \infty} \frac {1-\frac{t}{(t-1)}}{1-\sqrt{\frac{t}{(t-1)}}}[/tex]

derivative turns out to be:

[tex]\lim_{t \to \infty} \frac{\frac{1}{(t-1)^{2}}}{\frac{1}{2}(\frac{1}{(t-1)^{2}})(\frac{t}{t-1})^{-1/2}}[/tex]

and after the obvious term cancels we have:

[tex]\lim_{t \to \infty} \frac {1}{\frac{1}{2} (\frac{t}{t-1})^{-1/2}}[/tex]

and since it is a reciprocal of a reciprocal this is the same thing:

[tex]\lim_{t \to \infty} 2(\frac{t}{t-1})^{1/2}[/tex]

then with the limit applied:


[tex]\lim_{t \to \infty} 2(\frac{t}{t-1})^{1/2} = 2[/tex]

the step with the flipping of the reciprocals is really all I need check, I am fairly certain everything else is good.

thanks!
 
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Looks fine to me. Well done.
 
Make a substitution:

[tex] <br /> u = \sqrt{\frac{t}{t - 1}}<br /> [/tex]

and re-express your expression in terms of [itex]u[/itex]. When [itex]t \rightarrow \infty[/itex], what does [itex]u[/itex] tend to? You will have a simpler problem that can be solved by elementary methods (factorization).
 
I used a difference of squares and Dickfore's substitution to make it (1+u)(1-u) over (1-u). Why did you take the derivative? It just looks confusing?
 
Unit said:
I used a difference of squares and Dickfore's substitution to make it (1+u)(1-u) over (1-u). Why did you take the derivative? It just looks confusing?

Asphyxiated solved it by using l'Hopital on a 0/0 form. The given solution is just fine. It's simpler with a substitution, but that doesn't mean there is anything wrong with what Asphyxiated did.
 
Oh! Sorry! I meant no harm, and I certainly didn't mean to imply that there was anything wrong with Asphyxiated's solution. My post may have sounded rude but really I was just surprised. I had temporarily forgotten about l'Hopital's rule because we aren't allowed to use it in my grade 12 class, and when I saw this limit I just thought of difference of squares right away.
 
I did it the way that I did it because it is in the section of the book to learn to use l'hopital's rule, that is it.
 
Unit said:
Oh! Sorry! I meant no harm, and I certainly didn't mean to imply that there was anything wrong with Asphyxiated's solution. My post may have sounded rude but really I was just surprised. I had temporarily forgotten about l'Hopital's rule because we aren't allowed to use it in my grade 12 class, and when I saw this limit I just thought of difference of squares right away.

I didn't think you were trying to be rude or anything, don't apologize. But you asked "why the derivative", so I answered.
 

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