Very lost with l'Hospital's rule

  • Thread starter coverticus
  • Start date
  • Tags
    Lost
In summary, the conversation was about evaluating the limit of an expression as x approaches infinity and determining the type of indeterminate form it belongs to. The participants discussed using l'Hopital's rule and taking the logarithm to rearrange the expression into a 0/0 form. They also discussed the application of the chain rule and ultimately arrived at the answer of e^-1.
  • #1
coverticus
17
0

Homework Statement


Evaluate lim x[tex]\rightarrow[/tex] infinity of ([tex]\frac{x}{x+1}[/tex])[tex]^{}x[/tex], state explicitly the type of the indeterminate form.



Homework Equations





The Attempt at a Solution


I somewhat understand how to use l'Hospital's rule when the form is 0/0, but the inf/inf throws me off completely.
f'(x) = (x/(x+1))^x *ln(x/(x+1))
but from there or maybe the start I'm lost.
 
Physics news on Phys.org
  • #2
whether its 0/0 or inf/inf you proceed the same way...take the derivative of the top function and the derivative of the bottom function and evaluate the limit again
 
  • #3
First you need to change the expression to a 0/0 form. Hint: Take the log first, then try to rearrange that into a 0/0 form.
 
  • #4
I should have posted this sooner, but isn't it in the form 1^inf ?
 
  • #5
Yes, that's why you need to take the log first. Find the limit of the logarithm, then exponentiate to get the limit of the original expression.
 
  • #6
so you would get e^limx->inf of xln(x/(x+1)) and then what?
 
  • #7
Uh, rearrange xln(x/(x+1)) into a 0/0 form and apply l'Hopital to find the limit.
 
  • #8
ok I got that much, and I got x/(1/ln(x/(x+1))), but I'm lost as to evaluate that lim with l'Hospitals, would it be (1) / (1/ln(x/(x+1)))', if so I haven't the slightest as to how to get that derivative.
 
  • #9
It's a lot easier to write it as ln(x/(x+1))/(1/x). Use the chain rule to do the ln part.
 
  • #10
hey thanks for all your help guys, i got the answer (e^-1) finally and it was for a quiz so i appreciate it.
 

Related to Very lost with l'Hospital's rule

1. What is l'Hospital's rule?

L'Hospital's rule is a mathematical theorem that helps solve limits involving indeterminate forms, such as 0/0 or ∞/∞. It states that for a function f(x) and g(x) where both approach 0 or ∞ as x approaches a certain value, the limit of f(x)/g(x) is equal to the limit of f'(x)/g'(x), where f'(x) and g'(x) are the derivatives of f(x) and g(x) respectively.

2. When should I use l'Hospital's rule?

L'Hospital's rule should be used when evaluating a limit that results in an indeterminate form, such as 0/0 or ∞/∞. It is particularly useful when dealing with limits involving trigonometric, exponential, or logarithmic functions.

3. How do I apply l'Hospital's rule?

To apply l'Hospital's rule, first determine if the limit is in an indeterminate form. If it is, take the derivative of both the numerator and denominator separately. Then, evaluate the limit using the new derivatives. If the result is still an indeterminate form, continue taking derivatives until the limit can be evaluated.

4. Are there any restrictions when using l'Hospital's rule?

Yes, there are a few restrictions when using l'Hospital's rule. The functions in the limit must be differentiable in the neighborhood of the value being approached. Additionally, the limit must be in an indeterminate form, and both the numerator and denominator must approach 0 or ∞ as x approaches the same value.

5. Can l'Hospital's rule be used to solve any limit?

No, l'Hospital's rule can only be used to solve limits involving indeterminate forms. If a limit does not result in an indeterminate form, l'Hospital's rule cannot be applied. In these cases, other methods, such as direct substitution or algebraic manipulation, must be used to evaluate the limit.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
Replies
4
Views
2K
Replies
5
Views
1K
Replies
35
Views
3K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top