When Can i apply L'Hopital's rule?

  • Context: Undergrad 
  • Thread starter Thread starter zolit
  • Start date Start date
  • Tags Tags
    Apply L'hopital's rule
Click For Summary
SUMMARY

The discussion centers on the application of L'Hôpital's rule to prove that the derivative f'(0) exists and equals 0 for a function f that is differentiable on an interval containing 0, continuous at 0, and satisfies the limit condition lim f'(x) as x approaches 0 equals 0. Participants confirm that L'Hôpital's rule can be applied since both f and g(x) = x approach 0, creating a 0/0 indeterminate form. While L'Hôpital's rule is deemed valid for this problem, alternative methods such as Rolle's theorem are also considered, highlighting the importance of choosing the most effective approach.

PREREQUISITES
  • Understanding of L'Hôpital's rule and its conditions
  • Knowledge of limits and continuity in calculus
  • Familiarity with differentiable functions and their properties
  • Basic understanding of Rolle's theorem
NEXT STEPS
  • Study the conditions for applying L'Hôpital's rule in depth
  • Explore the implications of Rolle's theorem in limit problems
  • Practice solving limits that result in indeterminate forms
  • Investigate alternative methods for proving the existence of derivatives
USEFUL FOR

Students and educators in calculus, mathematicians focusing on analysis, and anyone looking to deepen their understanding of limits and derivatives in mathematical functions.

zolit
Messages
6
Reaction score
0
I am trying to work through the following problem:
if function is differentiable on an interval containing 0 except possibly at 0, and it is continuous at 0, and 0= f(0)= lim f ' (x) (as x approaches 0). Prove f'(0) exists and = 0.

I thought of using the definition of a limit to get to lim [ f(x)/x] then set g(x)=x and then use L'hopital's rule. The problem is - I'm not sure I can, is it enough to show that both f, g go to zero as x goes to zero to use it??

The alternative approach i was thinking of is considering two intervals (minus delta, 0) and (0, plus delta) and then using Rolle's theorem, but the solution seems to get too complicated from then.
 
Physics news on Phys.org
zolit said -- "I thought of using the definition of a limit to get to lim [ f(x)/x] then set g(x)=x and then use L'hopital's rule."

My point of view is this, when it comes to prove something, you can either based your solutions on definition or some existing theorem. L'hopital's rule requires that both f'(x) and g'(x) exist. And here in your problem you are asked to prove that f'(x) exists when x=0.

The approach I was thinking of is considering f'+(0) and f'-(0) both exist and are equal to each other.
 


L'Hopital's rule can be applied when we have an indeterminate form, such as 0/0 or infinity/infinity. In this case, it seems like you have correctly identified that the use of L'Hopital's rule is possible, as the limit you are trying to evaluate is in the form of 0/0.

To apply L'Hopital's rule, we need to have a differentiable function in both the numerator and denominator of the fraction. In this case, since f is differentiable on the interval containing 0, we can use L'Hopital's rule to evaluate the limit.

However, it is important to note that L'Hopital's rule is not always the best approach to solving a limit problem. In some cases, as you have mentioned, using other techniques such as Rolle's theorem may lead to a simpler solution. Ultimately, it is up to you as the problem solver to decide which method is most appropriate in each situation.

In this particular problem, it seems like using L'Hopital's rule is a valid approach. So, you can go ahead and use it to prove that f'(0) exists and is equal to 0. Just make sure to carefully check the conditions for applying L'Hopital's rule and to show all the necessary steps in your solution. Good luck!
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
922
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K