What are the conditions for applying L'Hopital's Rule?

  • Context: Undergrad 
  • Thread starter Thread starter Ad123q
  • Start date Start date
  • Tags Tags
    Conditions
Click For Summary
SUMMARY

The discussion centers on the conditions for applying L'Hôpital's Rule, specifically addressing the requirement that the derivative of the denominator, g'(x), must be non-zero near the point of interest, c. The user presents a case where both f(2) and g(2) equal zero, yet g'(x) equals zero at x=0.5, which is not equal to c. The conclusion drawn is that g'(x) only needs to be non-zero in an open interval around c, rather than strictly for all x ≠ c. This interpretation aligns with the definitions from various sources regarding the application of L'Hôpital's Rule.

PREREQUISITES
  • Understanding of limits and continuity in calculus.
  • Familiarity with differentiable functions and their derivatives.
  • Knowledge of L'Hôpital's Rule and its standard conditions.
  • Ability to compute limits using algebraic manipulation.
NEXT STEPS
  • Study the formal proof of L'Hôpital's Rule to understand its derivation.
  • Explore examples of applying L'Hôpital's Rule in various limit scenarios.
  • Learn about alternative methods for evaluating indeterminate forms, such as Taylor series.
  • Investigate the implications of the Mean Value Theorem in relation to L'Hôpital's Rule.
USEFUL FOR

Students of calculus, mathematics educators, and anyone seeking to deepen their understanding of limit evaluation techniques, particularly in the context of L'Hôpital's Rule.

Ad123q
Messages
19
Reaction score
0
Hi,

I'm slightly confused about one aspect of the conditions for applying L'Hopital's rule.

N.b. apologies in advance for the lack of LaTeX.

l'hospital's rule:
Let f,g:(a,b) → R be differentiable and let c ε (a,b) be such that f(c)=g(c)=0 and g'(x)≠0 for x≠c.
Then lim(x→ c)[f(x)/g(x)] = lim(x→ c)[f'(x)/g'(x)], provided latter limit exists.


I have lim(x→ 2)[(x²+x-6)/(x²-x-2)] = lim(x→ 2)[(2x+1)/(2x-1)]

I compute this to be 5/3 (which is correct) by L'Hopital's rule.
My point is that this function satisfies f(2)=g(2)=0, but does NOT satisfy g'(x)≠0 for x≠2, as the solution of g'(x)=2x-1=0 is x=0.5; i.e. g'(x)=0 for x=0.5, and 0.5≠c(=2).

So why are we able to apply L'Hopital in this case?
 
Physics news on Phys.org
If take (a,b)=(1,3), then the requirements are satisfied. The point is that you need g'(x) nonzero only "close" to c, i.e. on some open interval containing c.
 
I think you may have stated the conditions incorrectly. In particular, "g'(x)≠0 for x≠c", I believe, should be "g'(x)≠0 for x=c", since that is all that is necessary for the limit of f'(x)/g'(x) to exist (which seems to be consistent with the definitions I've checked from other sources, then if it still doesn't exist, you can apply it again, where it may eventually work).

I might be wrong though. Where did you get the definition of L'Hospital's Rule from?
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K