I love you, Professor Dave. I was teaching face-to-face classes with pencil-and-paper homework, just like back in the day, when suddenly, due to Covid-19, my classes are all online. I can make reasonably good math instructional videos, but not fast enough to keep up with four different courses in real time. I've been wandering UA-cam looking for good math videos to fit my learning objectives. This one is great. Thank you. --Professor Lisa.
A must watch Calculus series for all who are studying this subject. Clear presentation that includes organized content in a textbook format with intelligent, concise and step by step explanation of concepts and worked out examples. I enjoy learning from these videos. Thanks.
Thanks for the online lecture man! My teacher at school is so hopeless, she couldn't taught L'Hospital because there's no L'Hospital explanation on textbooks
i love when some other smarty has done the maths to explain something that makes sense logically. I remember working towards this in a calc class back as a group in the day, before we worked on this.
At 5:42 , there is an error. If you express -x² to its reciprocal, you should get -1/x². So the solution should look like: (1/x) × (-1/x²) = -1/x³ If you differentiate this three times, you should arrive at an answer of 0/6, which is coincidentally also 0
Hi Dave! I believe we cannot use L’Hôpitals rule for sin(theta)/theta, this is because we need to know the derivative of this in order to use l’hopital’s, thus we cannot use l’hopitals for us to find the derivative… Am I right? How can we calculate this derivatives without using L’Hôpital’s rule then? Great video!
You can't use L'Hopital's rule prove that lim (sin(x)/x) as x-->0 = 1. That's a circular argument because in order to prove that the derivative of sin(x) is cos(x) you need to know what value is lim (sin(x)/x) as x-->0.
What? Check out my tutorial on derivatives of trigonometric functions. You just plot the rate of change of sine and you get cosine. It couldn't be simpler.
@@ProfessorDaveExplains I was trying to say that we can't do a formal demonstration of the derivative of sin(x) using l'Hopital's rule. I didn't mean you did it on this presentation. btw, nice video and training exercices
Good. But is it logical to use L'Hopital's rule to find lim sin (x)/x, as we must know the limit before we differentiate sin(x)? There're so many similar examples.
What I don't get is this: in the first example of sin x/x, the limit here is the definition of the derivative of sin x at x=0. Then in the same line he writes down cos x as the derivative of sin x which he's not supposed to know as that is exactly the thing he is calculating. Isn't he making justified assumptions?
Question regarding the limit of x(ln x) = 0, the graph of that function shows there is no limit from the left side, which is why you approached the limit from the right side, but I thought the rule of limits state that a limit can only exist if the limit exists from both sides? There is the limit of peace-wise functions but that still requires the limit of both functions to exist from both sides. I am just assuming there is some limit rule I am missing. Would you be so kind as to elaborate on this?
Sorry Prof Dave but I evaluate your limit at 7:45 and my calculator says "Domain error" because Ln(0) gives domain error. How do you solve this? Thanks in advance.
0/infinity and infinity/0 will yield the limit to be 0 and infinity. An example of these function would be y=ln(x)/x and y=x/ln(x) as x goes to 0 from the right respectively. The limit would go to -infinity and 0.
So we can't use the rule for the infinity over (infinity - infinity) ? Or we have to make this form infinity over infinity and the solve by L'Hospital?
I don't understand the last exercise (ln x/x^2): When deriving once, I get (1/x)/2x. However, the 1/x then gets derived to - 1/x^2, and I don't see where the minus sign went in the answer? Altough I get that your answer must be right since the function is curving up and not down and so it makes sense that the second derivative is positive when x -> infinity. Can anyone explain?
Hi. Have seen recently an MIT video which shows a failure in L ´Hospital rule. Lim for x tending to infinite from (x + cos x) / x Thanks I admire your videos
Did you mean (x + sin x) / x? (x + cos x) / x does not apply, because it does not produce an indeterminate form. L'Hopital's rule didn't fail. The failure was attempting to use it when it doesn't apply.
For the question 1 at the end, why does limx->∞ 2x-9/6x+7 = limx->∞ 2/6 ? what happened to the -9 and +7 as I got to the part where i differentiate n wasnt sure what to do next. thank you for the great vid anyways
@JinkunYan The thing is, to rigorously prove that d/dx (sin (x)) = cos(x), you should actually use the limit definition, not just plot some points. While you're proving with the limit, you'll have to know the result of lim_(h->0) sin(h)/h Similarly, you'll also need to know lim_(h->0) (cos(h) - 1)/h The traditional way to prove these limits is to geometrically show that, for an angle θ near 0 1 >= sin(θ)/θ >= cos(θ) As θ approaches 0, the expression is "squeezed" by 1 on both sides, so it has to be 1 From here, the other limit is not that bad, you can use the pythagorean identity and use the result of this limit we've found So basically, when you use L'Hôpital's rule, you've to know already the derivative of sine, but to show what the derivative for sine is, you actually have to know the result for this limit already
There is no real life use, it's pure mathematics, but limits are the answer we get when we substitute the closest value to the variable, when we won't get a definete value when we substitute the exact variable.
I have confusion: You said: 0*(infinity) = indeterminate (but 0*any number=0 [right?]) and again (infinity)^0 = indeterminate (but anything raised to zero equals 1, right?) and again 1^(infinity) = indeterminate (but one raised to any number equals 1, right?) So what do you mean by that. In other words what kind of expressions do you call INDETERMINATE?
1:52 WRONG. Lim x->0 (sin x)/x can't be solved using L'Hopital's rule. Because the derivative of Sin x = Cos x is demonstrated by solving Lim x->0 (sin x)/x , the same that we are trying to solve, it means circular reasoning and that's invalid (and really bad). Zero points to an exam question if you do that. It's a shame you did that to an otherwise great video.
@@ProfessorDaveExplains You can conjecture a derivative graphically but can't prove it graphically. You prove a derivative from its definition, which is always a limit. Instead and incidentaly to my point, you need a graphical, or rather geometrical proof to find Lim x->0 (sin x)/x. So back to the original point, you can't formally use L'Hopital's rule to find the limit x->0 (sin x)/x because the derivative of sin x uses limit x->0 (sin x)/x to find cos x, hence the circular reasoning.
I love you, Professor Dave. I was teaching face-to-face classes with pencil-and-paper homework, just like back in the day, when suddenly, due to Covid-19, my classes are all online. I can make reasonably good math instructional videos, but not fast enough to keep up with four different courses in real time. I've been wandering UA-cam looking for good math videos to fit my learning objectives. This one is great. Thank you. --Professor Lisa.
A must watch Calculus series for all who are studying this subject. Clear presentation that includes organized content in a textbook format with intelligent, concise and step by step explanation of concepts and worked out examples. I enjoy learning from these videos. Thanks.
Your channel is absolutely amazing! Thank you for helping students everywhere!
Great video. You taught the rule in a way that saved me from going to the Hospital, lol.
Thanks for the online lecture man!
My teacher at school is so hopeless, she couldn't taught L'Hospital because there's no L'Hospital explanation on textbooks
lol, that would have sucked bad. I did had good lecture and note but the teacher pulled Eminem on us so hard to understand the lyrics
@@Lostwolf16 :')
Thanks a lot for your wonderful explanation..Now I am confident with this concept. 🇮🇳
Learning calculus for free is so enjoyable. Thanks Prof Dave!
This rule is just so neat!
Finally understood lopital rule after HOURS. Thank you professor!
Thanks Dave!
i love when some other smarty has done the maths to explain something that makes sense logically. I remember working towards this in a calc class back as a group in the day, before we worked on this.
thank you professor Dave i got these lecture after 4 years at the day you uploaded and helped me to understand l'hopital's rule
thank you again
Your all videos are so informative n easy to understand.n tomorrow is my exam 😅.
Writing this bcz these videos helped me a lot.love from india❤️.
Thx for everything ♥️
At 5:42 , there is an error. If you express -x² to its reciprocal, you should get -1/x².
So the solution should look like:
(1/x) × (-1/x²) = -1/x³
If you differentiate this three times, you should arrive at an answer of 0/6, which is coincidentally also 0
no words can describe how grateful I am prof ❤
Thanks love from India!!
Thank you sir for your dedication and for making this free! 🙏
Thank you Professor Dave, please may you talk about the origin and the statement of L'Hospital rule.
Thanks a lot for organizable, understandable and excellent explanation!!
No one can teach as like u sir😁😀😂😀👍👌
The best professor in the woooooorld we love youuuuu sir 🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹
Thank you so mcuh for the explaination! Was very easy to follow along and understand L'Hospital's rules
I m from India 🇮🇳🇮🇳watching your lecture your videos very simply explain me the topic thank you so much sir for your efforts 🙏🙏🙏🙏🙏
Thanks so mach
Thi is an important
Concept in
Learning
Limit❤
this was so helpful!
This man is a hero!
From your lecture I got to know many things. ❤
many many thanks
Thank youuu 🤗🤗
stupid girl
@@tse278 bruh moment 69420
Your explanation is very nice professor ☺️
I finally get why asymptote of the range of function is that!!!!
I am so happppppppppppyyyyy!!!!!!!!!
Hi Dave!
I believe we cannot use L’Hôpitals rule for sin(theta)/theta, this is because we need to know the derivative of this in order to use l’hopital’s, thus we cannot use l’hopitals for us to find the derivative… Am I right? How can we calculate this derivatives without using L’Hôpital’s rule then?
Great video!
@0:25 By sandwich theorem isn't it =1
I love you professor, you always get me out of trouble, hope I meet you one day to thank you f2f
i've definitely subscribed , it helps me more than it helps you for sure!!!
Thank You! It was great!
Thank you sir...!! Its easy to understand properly
Perfect teaching totally understood
this is absolutely awesome thanks man
You're the great sir dave
Thank you!💜
thanks
thankyou so much bro!
You can't use L'Hopital's rule prove that lim (sin(x)/x) as x-->0 = 1. That's a circular argument because in order to prove that the derivative of sin(x) is cos(x) you need to know what value is lim (sin(x)/x) as x-->0.
What? Check out my tutorial on derivatives of trigonometric functions. You just plot the rate of change of sine and you get cosine. It couldn't be simpler.
@@ProfessorDaveExplains I was trying to say that we can't do a formal demonstration of the derivative of sin(x) using l'Hopital's rule. I didn't mean you did it on this presentation. btw, nice video and training exercices
Could you go over indeterminate products please?
Many many thanks, sir !!
I am gonna recommend this playlist to all the suffering 11th grade friends i have xD
Good. But is it logical to use L'Hopital's rule to find lim sin (x)/x, as we must know the limit before we differentiate sin(x)? There're so many similar examples.
What I don't get is this: in the first example of sin x/x, the limit here is the definition of the derivative of sin x at x=0. Then in the same line he writes down cos x as the derivative of sin x which he's not supposed to know as that is exactly the thing he is calculating. Isn't he making justified assumptions?
look earlier in the calculus playlist for a tutorial on finding the derivatives of trigonometric functions, it's quite well derived
You save me from Hospital
Question regarding the limit of x(ln x) = 0, the graph of that function shows there is no limit from the left side, which is why you approached the limit from the right side, but I thought the rule of limits state that a limit can only exist if the limit exists from both sides? There is the limit of peace-wise functions but that still requires the limit of both functions to exist from both sides. I am just assuming there is some limit rule I am missing. Would you be so kind as to elaborate on this?
Sorry Prof Dave but I evaluate your limit at 7:45 and my calculator says "Domain error" because Ln(0) gives domain error. How do you solve this? Thanks in advance.
Same question
Oh got it now
Because it's 0+
Excellent
Thank you sir
Thanks sir your my hero
Which hospital is that?
It’s the one in France
The one you go to when you think about this too much and get an aneurysm
I have a question: How can we know which function we are inverting and placing under the other? Please answer me
Wow! I finally get it!!!
thanks again sir
love you dave
Professor thank u, From 🇮🇳 India😢
Hey Dave, I'm a bit confused
Why didn't you use the quotient rule to solve for limx->0 (e^x/x^3)
Thank you, hope to hear from you soon
Rule applies only for when you take derivatives of quotients not for limits. 💜
Sir if it is 0/infinity or infinity/zero,,would we apply the L.rule?
Topi Ado no sir
@@cristiansantos5070 girl#me
Topi Ado sorry!
0/infinity and infinity/0 will yield the limit to be 0 and infinity. An example of these function would be y=ln(x)/x and y=x/ln(x) as x goes to 0 from the right respectively. The limit would go to -infinity and 0.
So we can't use the rule for the infinity over (infinity - infinity) ?
Or we have to make this form infinity over infinity and the solve by L'Hospital?
Thank u sir!
YOU SAVED ME
I don't understand the last exercise (ln x/x^2): When deriving once, I get (1/x)/2x. However, the 1/x then gets derived to - 1/x^2, and I don't see where the minus sign went in the answer? Altough I get that your answer must be right since the function is curving up and not down and so it makes sense that the second derivative is positive when x -> infinity. Can anyone explain?
Hi. Have seen recently an MIT video which shows a failure in L ´Hospital rule. Lim for x tending to infinite from (x + cos x) / x Thanks
I admire your videos
Did you mean (x + sin x) / x?
(x + cos x) / x does not apply, because it does not produce an indeterminate form.
L'Hopital's rule didn't fail. The failure was attempting to use it when it doesn't apply.
Amazing
Super sir
love you sir
what about when x approache to zero for absolute value of x over x does the rule work?!
I am 10 years old and finish the whole calculus course smart ha
पागल हो क्या
For the question 1 at the end, why does limx->∞ 2x-9/6x+7 = limx->∞ 2/6 ? what happened to the -9 and +7 as I got to the part where i differentiate n wasnt sure what to do next. thank you for the great vid anyways
He took the derivative again
derivative of 2x-9/6x+7 = 2/6
4:28 I guess here L'hopital rule will also give correct answer .
Thank 🥰
2:15 Actually you can’t use L’Hop here
how so?
You need to give the reason to prove you are right, rather than said' you are wrong'
@JinkunYan The thing is, to rigorously prove that d/dx (sin (x)) = cos(x), you should actually use the limit definition, not just plot some points.
While you're proving with the limit, you'll have to know the result of
lim_(h->0) sin(h)/h
Similarly, you'll also need to know
lim_(h->0) (cos(h) - 1)/h
The traditional way to prove these limits is to geometrically show that, for an angle θ near 0
1 >= sin(θ)/θ >= cos(θ)
As θ approaches 0, the expression is "squeezed" by 1 on both sides, so it has to be 1
From here, the other limit is not that bad, you can use the pythagorean identity and use the result of this limit we've found
So basically, when you use L'Hôpital's rule, you've to know already the derivative of sine, but to show what the derivative for sine is, you actually have to know the result for this limit already
this channel def the best at explaining shit
Thank you jesus
Done.
I still don't get it
I got it after 6th time 🥹
It ain't that hard by solving problems you will understand is better. 😊
i loved it
I dont know why sinx derivative to cosx
Check out my tutorial on derivatives of trig functions, I derive them graphically.
what if g(x)=1 for example and we apply this, we get f'(x)/0 if defined then it's infinity. just guessing not sure, HELP
Lim (x->inf, (x+cos(x))/x ) is indeterminate, but L'hopital's rule fails.
You're awesome
very bad
Thank you Jesus
Bro wtf💀
i think the second comprehansion questoin is wrong because its ans is not maching mine i have done several times
i love infinity
How to define a limit using real life examples
there are no real life examples! this is mathematics.
There is no real life use, it's pure mathematics, but limits are the answer we get when we substitute the closest value to the variable, when we won't get a definete value when we substitute the exact variable.
It's important but really often more background for integrals and derivatives which 100% have real-world applications :)
I thought this was going to show why L'Hopital's rule works (graphically? I don't know).
Bro casually solves infinity 🎉
Great
How computer and human solve the no solution problem?
I have confusion:
You said:
0*(infinity) = indeterminate (but 0*any number=0 [right?])
and again (infinity)^0 = indeterminate (but anything raised to zero equals 1, right?)
and again 1^(infinity) = indeterminate (but one raised to any number equals 1, right?)
So what do you mean by that. In other words what kind of expressions do you call INDETERMINATE?
No. Infinity is not a number, so rules of arithmetic do not apply.
I wrote it.
just came to know how to pronounce this word !!
good sir.
i
1:52 WRONG. Lim x->0 (sin x)/x can't be solved using L'Hopital's rule. Because the derivative of Sin x = Cos x is demonstrated by solving Lim x->0 (sin x)/x , the same that we are trying to solve, it means circular reasoning and that's invalid (and really bad). Zero points to an exam question if you do that. It's a shame you did that to an otherwise great video.
What? You don't need limits to get the derivative of sin x.
@@ProfessorDaveExplains the very definition of ANY derivative is a limit.
@@GustavoMerchan79 Yeah but you can do it graphically.
@@ProfessorDaveExplains You can conjecture a derivative graphically but can't prove it graphically. You prove a derivative from its definition, which is always a limit. Instead and incidentaly to my point, you need a graphical, or rather geometrical proof to find Lim x->0 (sin x)/x. So back to the original point, you can't formally use L'Hopital's rule to find the limit x->0 (sin x)/x because the derivative of sin x uses limit x->0 (sin x)/x to find cos x, hence the circular reasoning.
The main problem there is an equal sign between Lim x->0 (sinx)/x and Lim x->0 cosx