0:10 This is L'opi Lu, NOT L'Hopital's rule 😂 It's missing the assumption that limf and limg=both 0 or both infinity. Counter example: f(x)=x^2 , g(x)=x. As x approaches 1, Lim(f'/g')=2, while lim(f/g)=1
I am sorry, but the hypotesis "f(x)/g(x) is an indeterminate form" is missing from yuor statement of the theorem. E.G., take f(x) = x and g(x) = x + 1 at 0.
What you stated was not Lhopitals rule. The original limit of the quotient MUST be equal to either the I determined form 0/0 or inf/inf so you can use the thesis of the theorem.
What if the limit of f'/g' gives you L=infinity ? Can you always use l'Hôpitals rule for any limit in this case? Or does it just work for special cases like a
this was very helpful! always wanted to see counterexamples like this. The hypothesis "there is an open interval I (except at a)..." is kind of strange to me. Does the interval I have to contain a? What does it mean for an interval to contain infinity? like (M, infinity)?
If a is finite, yes, the interval must contain a (though the function may not be defined or differentiable at a). If a = ±∞, than it is indeed as you wrote, the interval must be (-∞,b) or (b,+∞).
the assumption g’ 0 simply grants you can write f’/g’. otherwise you cannot even consider the fraction in neighborhood of the accumulation point (in this case infinity). let alone its limit. it is like saying sin(x)/sin(x) tends to 1 as x tends to infinity. which makes no sense unless you redefine sin(x) in pi k to be 0. but then it’s not differentiable.
I guess the formulation of rule isnt full(some may think that you can plug in everything you want) . As i know both functions should be 0 or inf in some point to use L'Hopitales rule, correct me of I am wrong
That's too dumb from you to say and racist as well. The example this guy used is not supposed to be solved with l'hopital cause the indeterminate form of the limit should be 0/0 or inf/inf
0:10 This is L'opi Lu, NOT L'Hopital's rule 😂
It's missing the assumption that limf and limg=both 0 or both infinity.
Counter example: f(x)=x^2 , g(x)=x. As x approaches 1, Lim(f'/g')=2, while lim(f/g)=1
Most certainly, Argentina's flag sun
Or any of the other forms reducible to 0/0... i think there are 7 such forms or so
Another example: f(x) = 1 + 2x, g(x) = 1 + x with lim (f'/g') = 2 while lim (f/g) = 1.
Another example is f(x) = 1, g(x) = x, as x approaches 0. We wind up with zero equaling infinity.
I am sorry, but the hypotesis "f(x)/g(x) is an indeterminate form" is missing from yuor statement of the theorem. E.G., take f(x) = x and g(x) = x + 1 at 0.
Childhood: Ruined
😂
@@bprpfast 18k subs, 1 like on your comment
What you stated was not Lhopitals rule. The original limit of the quotient MUST be equal to either the I determined form 0/0 or inf/inf so you can use the thesis of the theorem.
Exactly
Call an ambulance, we are going to the Hôpital :)
Lol
Today, the patient didn't live. :(
ah yeah 0/0, something very serious that needs to be taken care off. xD
" but it's far away from the infinity" 😂
Classic!!
Well done Dr. Peyam
Brought a good chuckle on a Saturday morning (coffee, but no caramel flan) ... thank you ...
Wow, so many details!👏👏
Great video
Starting was amazing...😂 dr πm
Very nice video :D!!!!!!!!!
To be at the level where 'oscillation junk' works in your argument 😎!
I knew something was off when we cancelled the cosines
Yeah most of the counterexamples I find are mostly in the exam centre worst place possible ,well they aren't counterexamples but a dead end
What about x^2/[sqrt(x^2+1)] or (e^x - e^(-x))/(e^x + e^(-x))
Where you get infinite loops. Which hypotheses does lhopital fail?
Dr. Peyam, great video!
I am wondering how this concept of an interval near infinity can be made more rigorous? (11:16)
Just say that there is no M > 0 such that for every b > M the interval (b, infinity) doesn't contain points x such that f(x) = 0.
Does lhopital fail if
a) the derivative is not continuous? Eg a limit involving (x^2)sin(1/x)
b) derivative does not exist?
Is Stolz-Cesàro Lu Wrong? R U Series?
What if the limit of f'/g' gives you L=infinity ? Can you always use l'Hôpitals rule for any limit in this case? Or does it just work for special cases like a
Interesting! I think it should be fine, but not sure
this was very helpful! always wanted to see counterexamples like this.
The hypothesis "there is an open interval I (except at a)..." is kind of strange to me. Does the interval I have to contain a? What does it mean for an interval to contain infinity? like (M, infinity)?
If a is finite, yes, the interval must contain a (though the function may not be defined or differentiable at a). If a = ±∞, than it is indeed as you wrote, the interval must be (-∞,b) or (b,+∞).
the assumption g’ 0 simply grants you can write f’/g’. otherwise you cannot even consider the fraction in neighborhood of the accumulation point (in this case infinity). let alone its limit. it is like saying sin(x)/sin(x) tends to 1 as x tends to infinity. which makes no sense unless you redefine sin(x) in pi k to be 0. but then it’s not differentiable.
I guess the formulation of rule isnt full(some may think that you can plug in everything you want) . As i know both functions should be 0 or inf in some point to use L'Hopitales rule, correct me of I am wrong
Two deers are also watching 1:43
Early gang! Yeah!
Nice👍
This Is sick
What's best book for abstract algebra? Dr Peyam reply me please
I don’t like abstract algebra
Cool 😎 video
Yeah ratios are weird.
0:0 no one wins.
0:1 only one side wins
1:1 50/50 on who wins
1:2 33/66 one side has an advantage
I m curious about his voice
👍
Never apply L'Hospital blindly
As always, you're a crack pal; when I grow up, I want to be like you. OOO
What's next?!! Stölz is WRONG???!!!
11:06
Peyam
Pi over 2 plus Peyam
Sitting in a tree
D-I-F-F-E-R-E-N-T-I-A-T-I-N-G
A moment before that happened I had some idea that he would do the thing. Yet I was surprised when he did the thing!
Lhospital was a french "matheux", not really reliable!
That's too dumb from you to say and racist as well. The example this guy used is not supposed to be solved with l'hopital cause the indeterminate form of the limit should be 0/0 or inf/inf