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## AP®︎ Calculus BC (2017 edition)

# Proof of special case of l'Hôpital's rule

AP.CALC:

LIM‑4 (EU)

, LIM‑4.A (LO)

, LIM‑4.A.1 (EK)

, LIM‑4.A.2 (EK)

L'Hôpital's rule helps us find limits in the form limit, start subscript, x, \to, c, end subscript, start fraction, u, left parenthesis, x, right parenthesis, divided by, v, left parenthesis, x, right parenthesis, end fraction where direct substitution ends in the indeterminate forms start fraction, 0, divided by, 0, end fraction or start fraction, infinity, divided by, infinity, end fraction.

The rule essentially says that

*if*the limit limit, start subscript, x, \to, c, end subscript, start fraction, u, prime, left parenthesis, x, right parenthesis, divided by, v, prime, left parenthesis, x, right parenthesis, end fraction*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?(8 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.(26 votes)

- Can I proof it when f(x) is infinity?(7 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(3 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)

- 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)(7 votes) - 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)|=∞.(3 votes)

- Sal said it was only for the special case. So what about the proof of the general case?(4 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)
- how does the lim x>0 ((ln(6x))^2/(ln(2x)^2)) is =1?(1 vote)
- limit of (ln(6x)^2/ln(2x)^2) as x->0

= (limit of ln(6x)/ln(2x)) ^ 2

= -∞/-∞

-∞/-∞ is indeterminate so we can apply L'Hospital's Rule

(limit of ln(6x)/ln(2x)) ^ 2

= (limit of (1/x)/(1/x)) ^ 2

= (limit of 1) ^ 2

= 1 ^ 2

= 1(1 vote)

- What is the proof for, if f(x) = ±∞ and g(x) = ±∞,

lim x-->c [f'(x)/g'(x)] = lim x-->c [f(x)/g(x)]?(1 vote) - 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?(1 vote)