A seemingly tricky limit - L'Hopital

In summary, the conversation was about finding the limit of the function \frac {1-\frac{t}{(t-1)}}{1-\sqrt{\frac{t}{(t-1)}}} as t approaches infinity. One person used a derivative to solve it, while another used a substitution and factoring. Both methods were valid and resulted in the same answer of 2.
  • #1
Asphyxiated
264
0

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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  • #2
Looks fine to me. Well done.
 
  • #3
Make a substitution:

[tex]

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

[/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).
 
  • #4
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
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.
 
  • #6
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
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
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.
 

1. What is L'Hopital's rule and when is it used?

L'Hopital's rule is a mathematical concept that is used to solve limits involving indeterminate forms, such as 0/0 or ∞/∞. It states that if the limit of the quotient of two functions is indeterminate, then the limit can be found by taking the derivative of both the numerator and denominator and then evaluating the limit again.

2. How do I know when to use L'Hopital's rule?

You can use L'Hopital's rule when you encounter a limit that is in an indeterminate form. Indeterminate forms occur when the limit cannot be easily evaluated by substituting the limit value into the function. Common indeterminate forms include 0/0, ∞/∞, and 1^∞.

3. Is L'Hopital's rule applicable to all limits?

No, L'Hopital's rule can only be used for certain types of limits. It is applicable to limits that are in an indeterminate form, but it cannot be used for all indeterminate forms. For example, it cannot be used for limits involving ∞ - ∞ or 0 x ∞.

4. Can L'Hopital's rule be applied multiple times?

Yes, L'Hopital's rule can be applied multiple times as long as the limit remains indeterminate. This means that if the limit still results in an indeterminate form after using L'Hopital's rule once, then it can be applied again.

5. Are there any limitations to L'Hopital's rule?

Yes, L'Hopital's rule has some limitations. It can only be used for functions that are differentiable, and it cannot be used for limits involving discontinuous functions. Additionally, it should be used with caution as it may not always provide the correct answer or may lead to incorrect results if used incorrectly.

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