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### Course: AP®︎ Calculus BC (2017 edition) > Unit 6

Lesson 1: L'Hôpital's rule# Proof of special case of l'Hôpital's rule

L'Hôpital's rule helps us find limits in the form $\underset{x\to c}{lim}{\displaystyle \frac{u(x)}{v(x)}}$ where direct substitution ends in the indeterminate forms $\frac{0}{0}$ or $\frac{\mathrm{\infty}}{\mathrm{\infty}}$ .

The rule essentially says that $\underset{x\to c}{lim}{\displaystyle \frac{{u}^{\prime}(x)}{{v}^{\prime}(x)}}$

*if*the limit*exists*, then the two limits are equal: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.

## Want to join the conversation?

- In the video of the proof: don't we need g'(a) to be nonzero in order to divide by it?(10 votes)
- 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.(30 votes)

- Can I proof it when f(x) is infinity?(8 votes)
- 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(9 votes)

- 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)(9 votes) - 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?(3 votes)
- 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.(10 votes)

- Sal said it was only for the special case. So what about the proof of the general case?(7 votes)
- Why is this a special case of L'Hôpital's Rule? Is this just the case where f(x) = 0 and g(x) = 0, or is there something else special about this case?(6 votes)
- 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.(1 vote)

- Can L'Hopital's rule be proved without assuming that f'(a) = 0 and g'(a) = 0 ?(3 votes)
- 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)|=∞.(4 votes)

- 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?(2 votes)
- yeah and also when they both equal 0(1 vote)

- at1:10, does f'(a)/g'(a) equal to lim x->a f'(x)/g'(x)? I think it's just direct substitution, right?(1 vote)
- 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.(2 votes)

- 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?(1 vote)