A seemingly tricky limit - L'Hopital

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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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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.