Proof of special case of l'H?pital's rule

L'H?pital's rule helps us find limits in the form

lim_{x \to c} \frac{u (x)}{v (x)}

where direct substitution ends in the indeterminate forms

\frac{0}{0}

\frac{\infty}{\infty}

The rule essentially says that if the limit

lim_{x \to c} \frac{u^{'} (x)}{v^{'} (x)}

exists, then the two limits are equal:

lim_{x \to c} \frac{u (x)}{v (x)} = lim_{x \to c} \frac{u^{'} (x)}{v^{'} (x)}

The AP Calculus course doesn't require knowing the proof of this fact, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn.

Khan Academy video wrapper

Proof of special case of l'H?pital's ruleSee video transcript

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Gabriela Asl? Rino Nesin
Posted 7 years ago. Direct link to Gabriela Asl? Rino Nesin's post “In the video of the proof...”
In the video of the proof: don't we need g'(a) to be nonzero in order to divide by it?
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(10 votes)
Answer
- 20leunge
  Posted 7 years ago. Direct link to 20leunge's post “Sorry about the confusion...”
  Sorry about the confusion earlier but here's my new answer: Yes, it has to be nonzero in order to divide it. If g'(a) is zero AND f'(a) is zero, then it is indeterminate form and you can just use L'Hopital's Rule again. If f'(a) is not zero, then the limit probably doesn't exist. I hope this answers your question.
  Comment on 20leunge's post “Sorry about the confusion...”
  (30 votes)
xyzPoKeFaNxyz
Posted 4 years ago. Direct link to xyzPoKeFaNxyz's post “Can I proof it when f(x) ...”
Can I proof it when f(x) is infinity?
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(8 votes)
Answer
- Yaaan
  Posted 4 years ago. Direct link to Yaaan's post “still let f(a)=0, g(a)=0....”
  still let f(a)=0, g(a)=0. but m(x) =1/g(x), n(x) = 1/f(x) So that m(a)= ∞, n(a) = ∞.
  Then m'(a)/n'(a) =
  lim(x==>a) {(m(x)-m(a))/(x-a)} / {(n(x)-n(a))/(x-a)} =
  lim(x==>a) {m(x)-m(a)} / {n(x)-n(a)} =
  lim(x==>a) {1/g(x)-1/g(a)} / {1/f(x)-1/f(a)} =
  lim(x==>a) {[g(a)-g(x)]/[(g(x)g(a)]} / {[f(a)-f(x)]/[(f(x)f(a)]}
  g(a)= 0, f(a) = 0 in numerator, so We can symplify it, then
  lim(x==>a) {[g(a)-g(x)]/[(g(x)g(a)]} / {[f(a)-f(x)]/[(f(x)f(a)]} =
  lim(x==>a) {[-g(x)]/[(g(x)g(a)]} / {[-f(x)]/[(f(x)f(a)]} =
  f(a)/g(a) = m(a)/n(a)
  then we proved m'(a)/n'(a)=m(a)/n(a),
  then lim(x==>a) m(x)/n(x) = m'(a)/n'(a) when m(a), n(a) are all infinite
  Comment on Yaaan's post “still let f(a)=0, g(a)=0....”
  (9 votes)
R Haq
Posted 4 years ago. Direct link to R Haq's post “Can't this qualify as a g...”
Can't this qualify as a general proof? Since l'hopitals rule only qualifies for indeterminate forms (0/0 and infinity/ infinity) if you've proven it for 0/0 you've infinity as well.
All indeterminate forms can be written as 0/0.
For example, consider a limit
lim_x->c f(x)/g(x)
which is infinity over infinity, one can equivalently say
lim_x->10^+ (1/g(x))/(1/f(x))
which is 0/0
and we've proven l'hopitals rule for 0/0
you can see more convertions between 0/0 and other forms like 1^infinity on the wikipedia page (https://en.wikipedia.org/wiki/Indeterminate_form)
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(9 votes)
Answer
abhi.devata
Posted 6 years ago. Direct link to abhi.devata's post “In the video, Sal works b...”
In the video, Sal works backwards from the formula that we use: lim(x==>c) of f(x)/g(x)=0/0 = lim(x==>c) of f'(x)/g'(x). But he starts the proof with the f'(x) and g'(x). That would mean that trial and error was performed. How do we prove this by starting with f(x) and g(x) instead of their derivatives?
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(3 votes)
Answer
- majumderzain
  Posted 6 years ago. Direct link to majumderzain's post “Here's an example with le...”
  Here's an example with less calculus:
  We want to prove that 4 = 16/8-7+9. Normally you would do:
  16/8-7+9 = 2-7+9 = -5+9 = 4. But you could also do this backwards:
  4 = -5+9 = 2-7+9 = 16/8-7+9.
  
  These proofs do the same thing, but one is more straightforward than the other. Sal chooses to start with lim(f'(x)/g'(x)) and get to lim(f(x)/g(x)) because the proof is more straightforward that way.
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  (10 votes)
Lee
Posted 5 years ago. Direct link to Lee's post “Sal said it was only for ...”
Sal said it was only for the special case. So what about the proof of the general case?
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(7 votes)
Answer
SULAGNA NANDI
Posted a year ago. Direct link to SULAGNA NANDI's post “Why is this a special cas...”
Why is this a special case of L'H?pital's Rule? Is this just the case where f(x) = 0 and g(x) = 0, or is there something else special about this case?
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(6 votes)
Answer
- Stefan Bonn
  Posted 6 months ago. Direct link to Stefan Bonn's post “Two things are different....”
  Two things are different. First and as you mentioned, this is only for f(x)=0 and g(x)=0. Second, it is assumed that f'(x) exists and g'(x) exists. This assumption is not made in the general form, where the limits of f'(x) and g'(x) are investigated.
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  (1 vote)
Taksh
Posted 6 years ago. Direct link to Taksh's post “Can L'Hopital's rule be p...”
Can L'Hopital's rule be proved without assuming that f'(a) = 0 and g'(a) = 0 ?
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(3 votes)
Answer
- Jakub Chora?z?y
  Posted 6 years ago. Direct link to Jakub Chora?z?y's post “Not really. You have to p...”
  Not really. You have to prove it for both cases; when f(a)=0 and g(a)=0, AND then when |f(a)|=∞ and |g(a)|=∞.
  Comment on Jakub Chora?z?y's post “Not really. You have to p...”
  (4 votes)
colinhill
Posted 10 months ago. Direct link to colinhill's post “Quick question, if this i...”
Quick question, if this is the special case of L'Hopital's Rule, what is the general case? Is it when f(a) and g(a) can also equal +/- infinity?
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(2 votes)
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- Forever Learner
  Posted 10 months ago. Direct link to Forever Learner's post “yeah and also when they b...”
  yeah and also when they both equal 0
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  (1 vote)
Liang
Posted a year ago. Direct link to Liang's post “at 1:10, does f'(a)/g'(a)...”
at
1:10
, does f'(a)/g'(a) equal to lim x->a f'(x)/g'(x)? I think it's just direct substitution, right?
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(1 vote)
Answer
- Venkata
  Posted a year ago. Direct link to Venkata's post “Yup! Note that this is on...”
  Yup! Note that this is only true if f'(x) and g'(x) are defined at x = a. But, as we've already established that, we can carry on with substituting the limit.
  Comment on Venkata's post “Yup! Note that this is on...”
  (2 votes)
Preslava Boycheva
Posted 5 years ago. Direct link to Preslava Boycheva's post “Since x is also approachi...”
Since x is also approaching 0 (it approaches a, which in turn approaches 0), shouldn't we also substitute f(x) and g(x) with 0 in the final equation of the limit of f'(x)/g'(x) as x approaches 0? How is this a proof when it actually takes us back to our initial problem with 0/0?
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(1 vote)
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