Log in Gabriela Aslı Rino Nesin 7 years agoPosted 7 years ago. Direct link to Gabriela Aslı 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? • (10 votes) 20leunge 7 years agoPosted 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. (30 votes) xyzPoKeFaNxyz 4 years agoPosted 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? • (8 votes) Yaaan 4 years agoPosted 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) = ∞. (9 votes) R Haq 4 years agoPosted 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. • (9 votes) abhi.devata 6 years agoPosted 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? • (3 votes) majumderzain 6 years agoPosted 6 years ago. Direct link to majumderzain's post “Here's an example with le...” Here's an example with less calculus: 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) Lee 5 years agoPosted 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? • SULAGNA NANDI a year agoPosted a year ago. Direct link to SULAGNA NANDI's post “Why is this a special cas...” 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) Stefan Bonn 5 months agoPosted 5 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. (1 vote) Taksh 6 years agoPosted 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 ? • (3 votes) Jakub Chorąży 6 years agoPosted 6 years ago. Direct link to Jakub Chorąży'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)|=∞. (4 votes) colinhill 9 months agoPosted 9 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? • (2 votes) Forever Learner 9 months agoPosted 9 months ago. Direct link to Forever Learner's post “yeah and also when they b...” yeah and also when they both equal 0 (1 vote) Liang a year agoPosted a year ago. Direct link to Liang's post “at 1:10, does f'(a)/g'(a)...” at • (1 vote) Venkata a year agoPosted 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. (2 votes) Preslava Boycheva 5 years agoPosted 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? • (1 vote)Want to join the conversation?
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
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)
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.
1:10
