I got maybe 7 minutes in, and this reminds me how I managed by some miracle to get through my freshman calculus year. Today I need strong coffee for this 20 minute video.
It may be useful to know that, when the denominator tends to ±∞, the de l'Hôpital rule always holds, no matter what the numerator tends to. Of course, it may not be an indeterminate form, but still lim (f/g) = lim (f'/g').
Hi, For fun: 3 "let's go ahead and", 1 "let's just go ahead and point out", 1 "now we want to go ahead and", 1 "I'll may be go ahead and", 1 "may be I'll go ahead and", 1 "the next thing that I want to notice", 2 "the next thing that I want to do", 1 "the important thing that I want to notice here", 2 "great".
11:38 Im trying to prove the second case where the limit of g(x) is negative infinity and I have a question. So we can change the proof to say that there is a delta2>0 such that 0
From what I gather, the only strict requirement is that g tend towards infinity. But if f doesn’t also tend towards infinity, then the limit will simply be zero and the theorem is unnecessary.
This seems harder than I remember. Maybe I'm misremembering, but I thought this proof was easier if you assume f(x)->0 and g(x)->0 instead of infinity.
I think the proof presented here is much more in depth than most people get when they first learn L'Hospital's Rule. I recall my professor sort of glossing over a lot of this and then "magically" coming up with how it all works out in the end.
SlimThrull well this is for an analysis course. The method of proof is much more important than the results, which is the opposite of a first year calculus class where most encounter this rule.
I got maybe 7 minutes in, and this reminds me how I managed by some miracle to get through my freshman calculus year. Today I need strong coffee for this 20 minute video.
In the definition of limit the neighborhood must be a perforated neighborhood so it is 0
It may be useful to know that, when the denominator tends to ±∞, the de l'Hôpital rule always holds, no matter what the numerator tends to. Of course, it may not be an indeterminate form, but still lim (f/g) = lim (f'/g').
Hi,
For fun:
3 "let's go ahead and",
1 "let's just go ahead and point out",
1 "now we want to go ahead and",
1 "I'll may be go ahead and",
1 "may be I'll go ahead and",
1 "the next thing that I want to notice",
2 "the next thing that I want to do",
1 "the important thing that I want to notice here",
2 "great".
good.
11:38 Im trying to prove the second case where the limit of g(x) is negative infinity and I have a question. So we can change the proof to say that there is a delta2>0 such that 0
Is he actually teaching a class with these videos and some of us are just weirdos tuning in for fun?
20:10
I love your username
the best subscriber
Good place to start at 0:00
5:40 notice that t could also be equal to b, where function is not necessarily defined either
Hi Michael, I think g(x) should greater than g(t) on 10:41. If g(x) is greater than or equal to g(t), it would be disastrous. What do you think?
I think you're right, everything seems to work without these two being equal to eachother.
Heckin good video
What a cool video. But when did you use the fact that the limit of f is tending to infinity?
From what I gather, the only strict requirement is that g tend towards infinity. But if f doesn’t also tend towards infinity, then the limit will simply be zero and the theorem is unnecessary.
@@dingo_dude Yeah you're right, it has to be an indeterminate form to be interesting. Thank you!
what is happening if g(x) approach negative inf, because I cannot yet understand what will happen in this case
Thank you so much for this great explanation 👍
ooh, time for a more rigorous refresher on everything that was brushed under the rug in AP calc
Good
Gotta love real analysis
Did you mean: L'Hôpital's
Spooky ghost goth girl on the blackboard :o
I dislike L'Hospital' theorem cause they did Bernoulli's bad even with taylor
This seems harder than I remember. Maybe I'm misremembering, but I thought this proof was easier if you assume f(x)->0 and g(x)->0 instead of infinity.
he did the 0/0 case in the last video, it's indeed way easier
@@metakaolin Oh okay. Then can't he prove this case by noting f(x)/g(x) is the same as (1/g(x))/(1/f(x))? Then just apply the 0/0 case
I think the proof presented here is much more in depth than most people get when they first learn L'Hospital's Rule. I recall my professor sort of glossing over a lot of this and then "magically" coming up with how it all works out in the end.
SlimThrull well this is for an analysis course. The method of proof is much more important than the results, which is the opposite of a first year calculus class where most encounter this rule.