# Homework Help: A seemingly tricky limit - L'Hopital

1. May 19, 2010

### Asphyxiated

1. The problem statement, all variables and given/known data

$$\lim_{t \to \infty} \frac {1-\frac{t}{(t-1)}}{1-\sqrt{\frac{t}{(t-1)}}}$$

2. Relevant equations

3. 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, heres what I did:

$$\lim_{t \to \infty} \frac {1-\frac{t}{(t-1)}}{1-\sqrt{\frac{t}{(t-1)}}}$$

derivative turns out to be:

$$\lim_{t \to \infty} \frac{\frac{1}{(t-1)^{2}}}{\frac{1}{2}(\frac{1}{(t-1)^{2}})(\frac{t}{t-1})^{-1/2}}$$

and after the obvious term cancels we have:

$$\lim_{t \to \infty} \frac {1}{\frac{1}{2} (\frac{t}{t-1})^{-1/2}}$$

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

$$\lim_{t \to \infty} 2(\frac{t}{t-1})^{1/2}$$

then with the limit applied:

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

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

thanks!

2. May 19, 2010

### Dick

Looks fine to me. Well done.

3. May 19, 2010

### Dickfore

Make a substitution:

$$u = \sqrt{\frac{t}{t - 1}}$$

and re-express your expression in terms of $u$. When $t \rightarrow \infty$, what does $u$ tend to? You will have a simpler problem that can be solved by elementary methods (factorization).

4. May 19, 2010

### Unit

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?

5. May 19, 2010

### Dick

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.

6. May 19, 2010

### Unit

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.

7. May 19, 2010

### Asphyxiated

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.

8. May 19, 2010

### Dick

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