1:37 The three conditions of continuity at a point x = c 5:10 Graphical representations of the three conditions of continuity 12:35 Finding whether or not three example functions are continuous at x=2 18:31 Statement for continuity of f(x) on an open interval (a,b) 19:35 Discussion of continuity of f(x) at the endpoints x=a and x=b 24:59 Proving f(x) = sqrt(16 - x^2) is continuous on [-4, 4] 35:38 Properties of limits f(x) and g(x) which are both continuous at x=c 38:20 Discussion of lim x→c f(x)/g(x) where g(c)=0 (holes & asymptotes) 40:38 Using continuity & limits to prove all polynomials are continuous everywhere 44:19 A rational function p(x)/q(x) is continuous everywhere except where q(x)=0, where we have either a hole or an asymptote 49:47 Finding discontinuities of f(x)=(x^2 - 4)/(x^2+x-6) 54:27 Proving f(x) = abs(x) is continuous everywhere 1:00:33 Proving we can separate limits via composition 1:05:02 Ex: Evaluating lim x→c |10 - 3x^2| 1:10:28 Ex: Finding inverse of f(x) = x^3 1:12:54 Concept of Intermediate Value Theorem 1:19:38 Approximating roots with IVT 1:21:27 Written rule for finding roots with IVT 1:23:05 Ex: estimating roots of y = x^3 - x - 1
The sad part about math is that your grade is often times more dependent on your teacher/professor than it is on your own intelligence. One teacher can make calc seem like quantum mechanics, while another can make the exact same subject as easy to understand as adding and subtracting. I am so lucky to have found you.
I'm not even in calculus yet and I legitimately enjoy watching this series/note taking. I learn so much from these lectures and know it will be very helpful down the road.
You are an AMAZING teacher! It's because of you that I can leave Calc 1 class completely confused and not stress over it because I know your videos will explain everything clearly when I get home. And best of all I can learn Calc 1 in my comfy lounging clothes while enjoying a nice snack :)
Seriously, thank you, Prof. Leonard! I've been trying for 3 weeks to teach myself Calc I for an online class. You have made it make sense finally. I can finally do my homework and breath. Thank you, thank you, thank you!
Am from Kenya taking a bachelor's degree in mathematics and I had given up and dread the calculus class till I met HIS EXCELLENCY PROFESSOR LEONARD.GOD BLESS YOU SIR.
For those of you who are seeking Lecture 1.3,you must understandably notice(If you really watched the entire Lecture 1.2) that Lecture 1.3 was combined to Lecture 1.2 because the last solved problem in Lecture1.2 was on the board in the beginning of Lecture 1.4.Hope that answers your question guys.
suraj awal Thanks, though I did get there, albeit four months later than you. Actually, your comment might have been more helpful if attached to Lecture 1.2 but, truly, thanks again anyway.
10:02 Just a note here: The limit does not exist if f(g)->infinity although you can see it that way. That's because infinity is not defined as a real number. I got what he meant, but that's mathematical incorrect. Anyway you write that the limit equals infinity although you should be aware that it's not a "limit value". Anyway this was a super pedagogical lecture! Thank you Professor Leonard!
The hardest part of calculus for me is how things can appear so obvious but require so much writing to prove the obvious at the same time. The |X| joke was on point about my feeling. It just seems so difficult having to wrap my head around writing these things down, I almost feel pressured I have to prove 2+2=4.
Yo, thought was the only one🤣🤣. He ain't that bad but you've got to pay so why despise the free UA-cam content from which he might have learnt from too. We should downvote the ad.
This is the best, you do not even have to force the stuff in, or crack your head because the explanations makes it crystal clear. Thank you Professor. Thanks.
18:31 Statement for continuity of f(x) on an open interval (a,b) 35:38 Properties of limits f(x) and g(x) which are both continuous at x=c 38:20 Discussion of lim x→c f(x)/g(x) where g(c)=0 (holes & asymptotes) 40:38 Using continuity & limits to prove all polynomials are continuous everywhere 44:19 A rational function p(x)/q(x) is continuous everywhere except where q(x)=0, where we have either a hole or an asymptote 1:12:54 Concept of Intermediate Value Theorem
:P my small calculus class is split with another class, so we've been watching these videos in absence of a full time teacher. Thanks for these, they're very helpful!
Amazing that this stuff was posted 11 years ago when I graduated. I never took calculus then, as I thought I would never need it in life. Here I am 11 years later going back to school and discovering just how incredibly powerful this subject is, and hoping to learn it well enough to tutor. Definitely inspired by this man's methods of teaching and aspiring to have such a great understanding of a very complex subject.
What I love is that Professor Leonard seems genuinely happy to share his love of math with others. I dont see this in many teachers. I belive many teachers start this way, but somehow the love is lost and the teaching becomes just a step of motions.
Wow, if you were my Calc prof, I'd have 0 problem giving you my undivided attention, I'd attend your class even if it was at 5am on a Sunday. Sorry, but it had to be said.
@@Crissybooable AND because he is hot - let's be honest about that. =P He even literally calls himself a "hot guy" in the video "Calculus 2 Lecture 8.1: Solving First Order Differential Equations By Separation of Variables" at 2:47:00, lmfao.
Really never ever learn calculus in this way.Loved it now l understand why all the thing are occurring and able to related question with graph. Epic teaching in calculus...😍😍
hallo professor, i really like Ur class, point to the point teaching methods. its really helpful. also for Ur kindness i started from bigging and i wish to go to the end. i wish it help me in the calculus 1 paper.very nice... NO BODY CAN DO WHAT U DID. THANKS A LOT.
For those of you who are seeking Lecture 1.3,you must understandably notice(If you really watched the entire Lecture 1.2) that Lecture 1.3 was combined to Lecture 1.2 because the last solved problem in Lecture1.2 was on the board in the beginning of Lecture 1.4.Hope that answers your question guys
+PANKAJ DAS He doesn't cover 1.3 in his class so there is no video for it. I believe he said something about the content of it being unnecessary for Calc 1
+musicluver7125 It's just combined, that's why the video is 3 hours long. 1.3 probably has to do with the squeeze theorem, which we didn't bother proving.
Professor Leonard, I just started a Calculus I class online. Your 1.1 - 1.3 videos helped me tremendously! I don't know if you will get this b/c the video is aged, but THANK YOU!!!
Professor Leonard is amazing. I wish I had him as my university math instructor. I have learned more in 5 days on these videos than I did in 5 months of college lectures. His style is ideal; energetic, comprehensive and precise. Does he do calc 2, 3, 4???
I definitely like your humor😇 Almost all my math teachers are stone-faced and work on how best they can beat the hell out of my back instead of how best they can drum the content into my head.
i dont take lectures of calc 1 anymore, because i know Sir Leonard will explain each and every thing that no one else can do. I only watch your lectures now.
This is a phenomenal supplementary, if not replacement, for my current Cal I class. Excellent teaching technique, the in depth look and useful examples really help make the information real and understandable beyond a plug and play equation game. Thanks for posting!
Dear professor, at 01:12:28 , the function is continuous both for f(x) = x^3 and its inverse between negative and positive infinity, but it is not the case for f(x) = x^2 (inverse domain is 0, +infinity), which is also a polynomial.
●[1:00:30]. Continuity of Composite Functions ◉[1:00:52]. Theorem: If the limit of 𝓖(x) as x approaches C exists and is equal to L, and 𝒇 is continuous at L, then the limit of 𝒇(𝓖(x)) as x approaches C is equal to 𝒇(L). ◉[1:02:17]. Demonstration: ○ The definition of the limit is used to express the limit of 𝓖(x) as x approaches C as L. ○ The continuity of 𝒇 at L is used to express 𝒇(L) as the limit of 𝒇(x) as x approaches L. ○ L is substituted with the limit of 𝓖(x) as x approaches C in the expression for 𝒇(L). ○ lim_(x → C) 𝒇(𝓖(x)) = 𝒇(L) = 𝒇( lim_(x → C) 𝓖(x) ) ◉[1:04:27]. Conclusion: The limit of a composite function can be *separated* into the limit of the inner function and the outer function: lim_(x → C) 𝓖(x) = L and 𝒇 is continuous at L => lim_(x → C) 𝒇(𝓖(x)) = 𝒇(L). ◉[1:04:56]. Example of How to Use Continuity of Composite Functions ○ [1:05:30]. Function: 𝒇(x) = |10 - 3x²|. ○ Procedure: ▹It is observed that the function is the composition of 𝓖(x) = 10 - 3x² and the absolute value function 𝒇(x) = |x|. ▹Since the absolute value function is continuous at all points, the theorem on continuity of composite functions can be applied. ○ [1:07:01]. Calculation of the Limit: ▹ The limit of 𝓖(x) as x approaches 4 is calculated: lim_(x → 4) (10 - 3x²) = -38. ▹ The absolute value function is applied to the result: |-38| = 38. ▹ Conclusion: limₓ→4 |10 - 3x²| = 38. ●[1:08:00]. Continuity of Inverse Functions ◉ [1:08:49]. Theorem: If 𝒇: Y -> Z and 𝓖: X -> Y are functions continuous at all points, then the composition 𝒇∘𝓖: X -> Z defined by (𝒇∘𝓖)(x) = 𝒇(𝓖(x)) is also continuous at all points. ◉ [1:08:49]. Theorem: If 𝒇 is continuous on its domain, then 𝒇⁻¹ is continuous on its domain, which is equal to the range of 𝒇. ◉ [1:10:40]. Example of Continuity of an Inverse Function ○ Function: 𝒇(x) = x³. ○ Continuity of 𝒇: Since 𝒇 is a polynomial, it is continuous on its entire domain, which is the set of all real numbers. ○ Range of 𝒇: The range of 𝒇 is also the set of all real numbers. ○ Inverse Function: 𝒇⁻¹(x) = ∛x. ○ Continuity of 𝒇⁻¹: According to the theorem on continuity of inverse functions, 𝒇⁻¹ is continuous on its domain, which is equal to the range of 𝒇, that is, the set of all real numbers. Intermediate Value Theorem ●[1:12:21]. Intermediate Value Theorem ◉ [1:12:40]. Introduction: The Intermediate Value Theorem is presented as an *application of continuity*. ◉ [1:13:00]. General Idea: If a function is continuous on a closed interval, and a value is taken between the values of the function at the endpoints of the interval, then there exists at least one point within the interval where the function takes that intermediate value. ◉ [1:15:43]. Formal Statement: ○ If 𝒇 is continuous on the closed interval [a, b], and k is a value between 𝒇(a) and 𝒇(b), then there exists *at least* one number c in the interval (a, b) such that 𝒇(c) = k. ●[1:17:35]. Interpretation of the Intermediate Value Theorem ◉ If a function is continuous on an interval, it cannot "jump" any value between the function values at the endpoints of the interval. ●[1:19:13]. Application of the Intermediate Value Theorem: Approximating Roots ◉ [1:19:20]. Introduction: The Intermediate Value Theorem can be used to approximate the roots of a function, that is, the points where the function crosses the x-axis. ◉ [1:19:38]. Procedure: ○ An interval [a, b] is sought where the function changes sign, that is, where 𝒇(a) and 𝒇(b) have opposite signs. ○ According to the Intermediate Value Theorem, if the function is continuous on [a, b] and 𝒇(a) and 𝒇(b) have *opposite signs*, then there is at least one root in the interval (a, b). ◉ [1:22:52]. Conclusion: If the signs of 𝒇(a) and 𝒇(b) are different, then there must be at least one root in the closed interval [a, b]. ●[1:23:25]. Example of How to Approximate Roots Using the Intermediate Value Theorem ◉ 𝒇(x) = x^3 - x - 1 ◉ [1:23:43]. Procedure: ○ An interval [a, b] is found where the function changes sign. In this example, the interval is used. ○ A table is created with x-values in the interval and the function is evaluated at each x-value. ○ It is observed where the function changes sign. In this example, the sign change occurs between x = 1.3 and x = 1.4 ○ The process is repeated with a smaller interval, in this case [1.3, 1.4], to obtain a more precise approximation of the root. ◉ [1:26:15]. Conclusion: The process can be repeated with increasingly smaller intervals to obtain an approximation of the root as precise as desired.
"removable discontinuity" means the whole is removable by changing the function to a piecewise function where the x,y value that fills the hole removes the discontinuity and the that function that has only one value in its domain is c.
Just so the advetisers know: No, I am not tired of searching UA-cam for math help. Yaymath, Kahn academy, and Professor Leonard are better than most things on TV these days.
Continuity ●[0:11]. Intuitive Definition of Continuity ◉ A function is continuous if it can be drawn without lifting the pencil from the paper. ◉ It has no holes, jumps, or asymptotes. ●[1:45]. Mathematical Definition of Continuity ◉ A function is continuous at a point C if it meets three conditions: (ⅰ). The function is defined at that point. 𝒇(c) is defined. (ⅱ). The limit of the function as x approaches C exists. Meaning the function approaches the same value from both sides. (ⅲ). The value of the function at point C is equal to the limit of the function as x approaches C. lim_(x → c) 𝒇(x) = 𝒇(c) ●[4:41]. Continuity in Terms of Drawing a Function ◉ If you trace a continuous function, you reach a point, fill in the point, and continue without interruptions. ●[5:35]. Graphical Examples of Continuity and Discontinuity ◉[5:37]. Example 1 (ⅰ): A Graph with a Hole at x = C ○ The function is not defined at x = C, so it is not continuous at that point. ○ The limit exists at x = C, but it is not equal to the value of the function (which does not exist). ○ This is an *removeable discontinuity*, as the hole can be "filled in" with a single point. ◉[7:30]. Example 2 (ⅱ): A Graph with a Jump at x = C ○ The function is defined at x = C. ○ The limit does not exist at x = C because the function approaches different values from the left and right. ○ This is a *jump discontinuity*. ◉[9:20]. Example 3 (ⅲ): A Graph with a Vertical Asymptote at x = C ○ The function is or not defined at x = C. ○ The limit exists at x = C and is infinite. ○ This is an *infinite discontinuity*. ◉[11:09]. Summary of Types of Discontinuity ○ There are three main types of discontinuities: 1. Discontinuities with asymptotes. 2. Jump discontinuities. 3. Removeable discontinuities (holes). ○ Removable discontinuities can be "filled in" with a single point. ○ Rational functions typically have removeable discontinuities or asymptotes. ○ Generally, when we encounter a jump discontinuity, the function will be a piecewise function. ●[13:07]Examples, Are these continuous at x = 2?: ◉[13:07]. 𝒇(x) = (x^2 - 4)/(x - 2) ◉[16:20]. 𝓖(x) = { (x^2 - 4) / (x - 2) if x ≠ 2 3 if x = 2 } ◉[17:36]. 𝒉(x) = { (x^2 - 4)/(x - 2) if x ≠ 2 4 if x = 2 } ●[18:25]. Continuity on an Interval ◉ If a function is continuous at every point between a and b, it is said to be continuous on the open interval (a, b). ●[19:35]. Continuity at the Endpoints of an Interval ◉ To determine if a function is continuous on a closed interval [a, b], you must verify continuity on the open interval (a, b) and continuity at the endpoints a and b. ◉ At the endpoints of an interval, one-sided limits are used to verify continuity. ◉ For a function to be continuous at an endpoint, the one-sided limit must exist and be equal to the value of the function at that endpoint. ●[22:50]. Mathematical Definition of One-Sided Continuity ◉ Continuity from the left: The limit of f(x) as x approaches C from the left must equal f(C). ○ lim_(x → c⁻) 𝒇(x) = 𝒇(c) ◉ Continuity from the right: The limit of f(x) as x approaches C from the right must equal f(C). ○ lim_(x → c⁺) 𝒇(x) = 𝒇(c) ●[25:41]. Function: 𝒇(x) = √(16 - x²) on the Closed Interval [-4, 4]. ◉Steps for the Demonstration: 1. Verify continuity on the open interval (-4, 4). 2. Verify continuity from the left at -4. 3. Verify continuity from the right at 4. ◉[29:11]. 1. Demonstration of Continuity on the Open Interval (-4, 4) ○ It must be shown that the limit of f(x) as x approaches C is equal to f(C) for any value of C between -4 and 4. ►lim_(x → c) 𝒇(x) = 𝒇(c) ◉[32:50]. Demonstration of One-Sided Continuity at the Endpoints ◉[34:53]. Conclusion of the Demonstration ● It has been shown that the function is continuous on the open interval and at the endpoints. ● Therefore, the function is continuous on the closed interval [-4, 4]. ●[36:00]. Properties of Continuity in Operations with Functions ◉ If f and g are continuous at a point C, then: ○ 𝒇 + 𝓖 is continuous at C. ○ 𝒇 - 𝓖 is continuous at C. ○ 𝒇 * 𝓖 is continuous at C. ◉ 𝒇 / 𝓖 is continuous at C unless 𝓖(C) = 0. ●[38:20]. Discontinuities in Rational Functions ◉ If 𝓖(C) = 0 in a rational function 𝒇 / 𝓖, then there is a discontinuity at C. ◉ The discontinuity can be a hole or an asymptote. ●[40:50]. Continuity of Polynomials ◉ Recall (ⅲ): ○ lim_(x → c) 𝒇(x) = 𝒇(c) ⇔ 𝒇 is continuous at c ◉ Recall the property of polynomial limits: the limit of a polynomial as x approaches any point can be evaluated simply by substituting x with that point. ○ ∀c ∈ ℝ: lim_(x → c) P(x) = P(c) ◉ This implies that *all polynomials are continuous at all points*. ○ ∀c ∈ ℝ: lim_(x → c) P(x) = P(c) ⇔ P is continuous at every point in ℝ ●[44:10]. Continuity of Rational Functions P(x) / Q(x) ◉ Combining the continuity property of polynomials with the continuity property of function division, it is concluded that *every rational function is continuous at all points except where the denominator [Q(x)] is zero*. At that point you have a *Discontinuity*. ○ [47:15]. Holes and Asymptotes in Rational Functions as Discontinuities. ⑴ Hole (0 / 0): In a rational function, if a common factor can be canceled by factoring the numerator and the denominator, the discontinuity corresponding to that factor is a hole. ⑵ Asymptote (constant / 0): If a factor of the denominator cannot be canceled, the discontinuity corresponding to that factor is an asymptote. ◉[50:07]. Example of How to Find Discontinuities in a Rational Function ○ Function: 𝒇(x) = (x² - 4) / (x² - x - 6). ○ Procedure: Discontinuities are found by setting the denominator to zero and solving the equation. ▹ Discontinuities: x = -2 and x = 3. ▹ Factor the function: 𝒇(x) = (x + 2)(x - 2) / (x + 3)(x - 2). ▹ It is observed that the factor (x - 2) cancels out, indicating a hole at x = 2. ▹ The factor (x + 3) does not cancel out, indicating an asymptote at x = 3. ●[55:00]. Demonstration of Continuity of the Absolute Value Function Procedure: The function is defined piecewise to avoid issues with limits at the endpoints. ◉ Continuity on Open Intervals: ○ For x > 0, the function is simply 𝒇(x) = x, which is a polynomial and therefore continuous. ○ For x < 0, the function is 𝒇(x) = -x, which is also a polynomial and therefore continuous. ◉ Continuity at x = 0: ○ The important thing here is to verify the definition of continuity at x=0; lim_(x → 0) |x| = |0| ○ Continuity from the left: lim_(x → 0⁻) |x| = lim_(x → 0⁻) -x = 0 ○ Continuity from the right: lim_(x → 0⁺) |x| = lim_(x → 0⁻) x = 0
I just love Professor's videos and explanations but I can't wrap my head around this. Can anyone explain at 1:04:00 professor made a jump from lim x tends to c f(g(x)) to f(L) and explains that if x is close to c, then g(x) must be equal to L. But, we're supposed to find the limit of f first, not the nested function g. If we're first calculating the limit of g first then f, and then proving that lim x->c f(g(x) = f(Lim x->c g(x)) then isn't the whole proof pointless. Because we're saying that we can put the limit of the composite function inside the nested function but we're literally taking the proof and using it in our steps.
After watching this I just realized that the limit properties directly correlate to the rules for finding derivatives. Can't believe that I learned everything I needed for Calculus in my first week of class without realizing it.
I really like your teaching style! Are you planning to make some videos about differential equations or linear algebra? Thanks for posting these videos about calculus.
Prof. Leonard, please collaborate with Khan Academy. I will be an asset, I'm sure and that way millions more can benefit from what you offer here. UA-cam is banned in a lot of countries. P.S why are your students always seem half asleep in your class. I wonder its because they know that they can watch the lectures later when they get home so they don't stay as alert.
This guy is bar none the best mathematics instructor I have seen on UA-cam. Whatever they are paying him where he teaches they need to double it.
I wish he taught at ASU :(.
So true XD
Agreed
This man saved me a lot of trouble for 2 semesters, this man is a legend❤😢
1:37 The three conditions of continuity at a point x = c
5:10 Graphical representations of the three conditions of continuity
12:35 Finding whether or not three example functions are continuous at x=2
18:31 Statement for continuity of f(x) on an open interval (a,b)
19:35 Discussion of continuity of f(x) at the endpoints x=a and x=b
24:59 Proving f(x) = sqrt(16 - x^2) is continuous on [-4, 4]
35:38 Properties of limits f(x) and g(x) which are both continuous at x=c
38:20 Discussion of lim x→c f(x)/g(x) where g(c)=0 (holes & asymptotes)
40:38 Using continuity & limits to prove all polynomials are continuous everywhere
44:19 A rational function p(x)/q(x) is continuous everywhere except where q(x)=0, where we have either a hole or an asymptote
49:47 Finding discontinuities of f(x)=(x^2 - 4)/(x^2+x-6)
54:27 Proving f(x) = abs(x) is continuous everywhere
1:00:33 Proving we can separate limits via composition
1:05:02 Ex: Evaluating lim x→c |10 - 3x^2|
1:10:28 Ex: Finding inverse of f(x) = x^3
1:12:54 Concept of Intermediate Value Theorem
1:19:38 Approximating roots with IVT
1:21:27 Written rule for finding roots with IVT
1:23:05 Ex: estimating roots of y = x^3 - x - 1
'Thank you'
omg you owe us a lot thaaankssss
Bless your soul
Citizens Educated god bless you 🙏
Thanks for the above
After more than a decade ,professor Leonard's lectures are still helping a lot of students
He takes the time to review prior algebra/trig topics and reiterates things to really help you recall and hammer into your head in an easy manner.
For those of you wondering, no, there is no lecture 1.3.
He took up directly from where he left in lecture 1.2.
By the way, thankyou sir.
thanks lol
you're brilliant thanks
This helped calm the panic
I swear I keep looking for it 😅🤧🙏🏽
Thank you.
The sad part about math is that your grade is often times more dependent on your teacher/professor than it is on your own intelligence. One teacher can make calc seem like quantum mechanics, while another can make the exact same subject as easy to understand as adding and subtracting. I am so lucky to have found you.
i was so in to the lecture @ 18:26 i said bless you when ever that kid sneezed and the whole library gave me that look.
I watch his videos on the treadmill and I'm drawing the functions with my hand I'm sure people think I'm crazy at the gym lol.
+Dan Morgan that multi tasking though. id trip on my face if i did that
where did dota come from?
DOTA is life.
@@gbb666-g7g agree
I'm not even in calculus yet and I legitimately enjoy watching this series/note taking. I learn so much from these lectures and know it will be very helpful down the road.
You're the best man... your videos are helping even after decades ❤🎉
You are an AMAZING teacher! It's because of you that I can leave Calc 1 class completely confused and not stress over it because I know your videos will explain everything clearly when I get home. And best of all I can learn Calc 1 in my comfy lounging clothes while enjoying a nice snack :)
right!
Lol the same feeling
@@mrcoffee315 Mr. Coffee haha
Seriously, thank you, Prof. Leonard! I've been trying for 3 weeks to teach myself Calc I for an online class. You have made it make sense finally. I can finally do my homework and breath. Thank you, thank you, thank you!
but girls r bad at math tho
This guy is definitely the best I have come across in my time on youtube. His explanations are the most complete I have ever seen.
Am from Kenya taking a bachelor's degree in mathematics and I had given up and dread the calculus class till I met HIS EXCELLENCY PROFESSOR LEONARD.GOD BLESS YOU SIR.
For those of you who are seeking Lecture 1.3,you must understandably notice(If you really watched the entire Lecture 1.2) that Lecture 1.3 was combined to Lecture 1.2 because the last solved problem in Lecture1.2 was on the board in the beginning of Lecture 1.4.Hope that answers your question guys.
thank you very much!
RADman305 You are welcome
suraj awal
Thanks, though I did get there, albeit four months later than you. Actually, your comment might have been more helpful if attached to Lecture 1.2 but, truly, thanks again anyway.
Yeah, I was looking at the stuff on the right hand side and saying hmm... that looks awfully familiar!
thanks
I love looking at these videos because since I'm self studying AP Calc AB, these lectures assure that I don't miss anything valuable :)
Back when someone could sneeze in class and no one would think anything of it lol
he called them sneeze grenades, which they are now lol
It’s so wholesome how everyone said Bless you 😊
I know it has been years since this was posted, but it just saved my life. Thankyou so much professor leonard!!!!
10:02 Just a note here: The limit does not exist if f(g)->infinity although you can see it that way. That's because infinity is not defined as a real number. I got what he meant, but that's mathematical incorrect. Anyway you write that the limit equals infinity although you should be aware that it's not a "limit value". Anyway this was a super pedagogical lecture! Thank you Professor Leonard!
@alin636 is correct.
He is definitely the best professor I've seen in my whole life. No words can describe how much he's awesome and Thank you in advance.
The hardest part of calculus for me is how things can appear so obvious but require so much writing to prove the obvious at the same time. The |X| joke was on point about my feeling. It just seems so difficult having to wrap my head around writing these things down, I almost feel pressured I have to prove 2+2=4.
this playlist is a gem. im so lucky i have found this. im gonna rock my next exam for sure
Yo, I hate seeing those "That Tutor Guy" ads whenever I watch anything involving math.
The videos from 2012 are helping.
Yo, thought was the only one🤣🤣. He ain't that bad but you've got to pay so why despise the free UA-cam content from which he might have learnt from too. We should downvote the ad.
@@winslowleach1835 Certainly in agreement with you.
This is the best, you do not even have to force the stuff in, or crack your head because the explanations makes it crystal clear. Thank you Professor. Thanks.
18:31 Statement for continuity of f(x) on an open interval (a,b)
35:38 Properties of limits f(x) and g(x) which are both continuous at x=c
38:20 Discussion of lim x→c f(x)/g(x) where g(c)=0 (holes & asymptotes)
40:38 Using continuity & limits to prove all polynomials are continuous everywhere
44:19 A rational function p(x)/q(x) is continuous everywhere except where q(x)=0, where we have either a hole or an asymptote
1:12:54 Concept of Intermediate Value Theorem
:P my small calculus class is split with another class, so we've been watching these videos in absence of a full time teacher. Thanks for these, they're very helpful!
Amazing that this stuff was posted 11 years ago when I graduated. I never took calculus then, as I thought I would never need it in life. Here I am 11 years later going back to school and discovering just how incredibly powerful this subject is, and hoping to learn it well enough to tutor. Definitely inspired by this man's methods of teaching and aspiring to have such a great understanding of a very complex subject.
What I love is that Professor Leonard seems genuinely happy to share his love of math with others. I dont see this in many teachers. I belive many teachers start this way, but somehow the love is lost and the teaching becomes just a step of motions.
40:28 "were gonna talk about that in 3.5 mins roughly...no, not roughly, exactly" LOL
Wow, if you were my Calc prof, I'd have 0 problem giving you my undivided attention, I'd attend your class even if it was at 5am on a Sunday. Sorry, but it had to be said.
because he is such a great teacher yes :)
Or maybe because he's an awesome human being, and the best maths teacher ever, and genuinly funny... and yes, good looking... but, so what?
He IS the total package..... damn him!
EXTRA T H I C C He can be my math Superman any day 😂
@@Crissybooable AND because he is hot - let's be honest about that. =P
He even literally calls himself a "hot guy" in the video "Calculus 2 Lecture 8.1: Solving First Order Differential Equations By Separation of Variables" at 2:47:00, lmfao.
Really never ever learn calculus in this way.Loved it now l understand why all the thing are occurring and able to related question with graph. Epic teaching in calculus...😍😍
hallo professor, i really like Ur class, point to the point teaching methods. its really helpful.
also for Ur kindness i started from bigging and i wish to go to the end. i wish it help me in the calculus 1 paper.very nice... NO BODY CAN DO WHAT U DID. THANKS A LOT.
Best teacher ever !!! im about to cry , hello form 2021
Professor Leonard is hecka better than my teacher, lol
Juan Rosa. Better looking, too 😂
For those of you who are seeking Lecture 1.3,you must understandably notice(If you really watched the entire Lecture 1.2) that Lecture 1.3 was combined to Lecture 1.2 because the last solved problem in Lecture1.2 was on the board in the beginning of Lecture 1.4.Hope that answers your question guys
+PANKAJ DAS are you presently taking a calculus class?
+PANKAJ DAS He doesn't cover 1.3 in his class so there is no video for it. I believe he said something about the content of it being unnecessary for Calc 1
+musicluver7125 Where did you see that at? The he said 1.3 is unnecessary because I didn't hear him say that.
+musicluver7125 It's just combined, that's why the video is 3 hours long. 1.3 probably has to do with the squeeze theorem, which we didn't bother proving.
+Ushikawa Really dude read my last comment it's not combined.
i want him as my professor . He is so good at explaining things and making it easier . i have so much respect for him :)
The person sneezing at 29:36 just reminded me that gone are the days you sneeze and you get a bless you
Thank you prof.Leonard.You are an awesome teacher and your lectures are so much fun........I understood every bit of it.
You are toooo amazing!!!!! 100000000000000000000000000000000 TIMESS BETTER THAN MY PROFESSORRRRRR!
Love your way of teaching sir...very few times i listen to teacher in class but now I am watching you and i am lov in it❤️
Professor Leonard,
I just started a Calculus I class online. Your 1.1 - 1.3 videos helped me tremendously! I don't know if you will get this b/c the video is aged, but THANK YOU!!!
Do you have a lecture on espilon-delta proofs? Your lectures are awesome btw!
The one guy to make me actually pursue math, amazing classes btw
You are the best teacher l have ever seen
Professor Leonard is amazing. I wish I had him as my university math instructor. I have learned more in 5 days on these videos than I did in 5 months of college lectures. His style is ideal; energetic, comprehensive and precise. Does he do calc 2, 3, 4???
Matthew Herron He has playlists for calc 2 & 3, I think. You could probably look on his channel to watch more!
you are one of the best teacher and bestest maths teacher I have ever studied
I definitely like your humor😇
Almost all my math teachers are stone-faced and work on how best they can beat the hell out of my back instead of how best they can drum the content into my head.
Good sneezes 29:35, 1:19:32
Thanks a bunch!
Clear, engaging, to the point. Keep doing what you're doing!
i dont take lectures of calc 1 anymore, because i know Sir Leonard will explain each and every thing that no one else can do. I only watch your lectures now.
CAN'T THANK YOU ENOUGH. YOU ARE THE BEST!!
I honestly require every College to interpret your enthusiasm, for not only your lectures but for the subject itself.
You are the BEST MATHEMATICIAN I´ve ever seen. I wish you were in Germany... Your students are soooo lucky......
I really really appreciate that you've shared this videos prof. thankyou very much
This is a phenomenal supplementary, if not replacement, for my current Cal I class. Excellent teaching technique, the in depth look and useful examples really help make the information real and understandable beyond a plug and play equation game. Thanks for posting!
12:03 Mann i could imagine the students' faces when the one guy asked the question haha
Dear professor, at 01:12:28 , the function is continuous both for f(x) = x^3 and its inverse between negative and positive infinity, but it is not the case for f(x) = x^2 (inverse domain is 0, +infinity), which is also a polynomial.
Really helpful, anytime I drop my riffle you keep me busy refreshing my cal
●[1:00:30]. Continuity of Composite Functions
◉[1:00:52]. Theorem: If the limit of 𝓖(x) as x approaches C exists and is equal to L, and 𝒇 is continuous at L,
then the limit of 𝒇(𝓖(x)) as x approaches C is equal to 𝒇(L).
◉[1:02:17]. Demonstration:
○ The definition of the limit is used to express the limit of 𝓖(x) as x approaches C as L.
○ The continuity of 𝒇 at L is used to express 𝒇(L) as the limit of 𝒇(x) as x approaches L.
○ L is substituted with the limit of 𝓖(x) as x approaches C in the expression for 𝒇(L).
○ lim_(x → C) 𝒇(𝓖(x)) = 𝒇(L) = 𝒇( lim_(x → C) 𝓖(x) )
◉[1:04:27]. Conclusion: The limit of a composite function can be *separated* into the limit of the inner function
and the outer function:
lim_(x → C) 𝓖(x) = L and 𝒇 is continuous at L => lim_(x → C) 𝒇(𝓖(x)) = 𝒇(L).
◉[1:04:56]. Example of How to Use Continuity of Composite Functions
○ [1:05:30]. Function: 𝒇(x) = |10 - 3x²|.
○ Procedure:
▹It is observed that the function is the composition of 𝓖(x) = 10 - 3x² and the absolute value
function 𝒇(x) = |x|.
▹Since the absolute value function is continuous at all points, the theorem on continuity of
composite functions can be applied.
○ [1:07:01]. Calculation of the Limit:
▹ The limit of 𝓖(x) as x approaches 4 is calculated: lim_(x → 4) (10 - 3x²) = -38.
▹ The absolute value function is applied to the result: |-38| = 38.
▹ Conclusion: limₓ→4 |10 - 3x²| = 38.
●[1:08:00]. Continuity of Inverse Functions
◉ [1:08:49]. Theorem: If 𝒇: Y -> Z and 𝓖: X -> Y are functions continuous at all points, then the composition
𝒇∘𝓖: X -> Z defined by (𝒇∘𝓖)(x) = 𝒇(𝓖(x)) is also continuous at all points.
◉ [1:08:49]. Theorem: If 𝒇 is continuous on its domain, then 𝒇⁻¹ is continuous on its domain, which is equal
to the range of 𝒇.
◉ [1:10:40]. Example of Continuity of an Inverse Function
○ Function: 𝒇(x) = x³.
○ Continuity of 𝒇: Since 𝒇 is a polynomial, it is continuous on its entire domain, which is
the set of all real numbers.
○ Range of 𝒇: The range of 𝒇 is also the set of all real numbers.
○ Inverse Function: 𝒇⁻¹(x) = ∛x.
○ Continuity of 𝒇⁻¹: According to the theorem on continuity of inverse functions, 𝒇⁻¹ is
continuous on its domain, which is equal to the range of 𝒇, that is, the set of all real numbers.
Intermediate Value Theorem
●[1:12:21]. Intermediate Value Theorem
◉ [1:12:40]. Introduction: The Intermediate Value Theorem is presented as an *application of continuity*.
◉ [1:13:00]. General Idea: If a function is continuous on a closed interval, and a value is taken between the
values of the function at the endpoints of the interval, then there exists at least one point
within the interval where the function takes that intermediate value.
◉ [1:15:43]. Formal Statement:
○ If 𝒇 is continuous on the closed interval [a, b], and k is a value between 𝒇(a) and 𝒇(b), then
there exists *at least* one number c in the interval (a, b) such that 𝒇(c) = k.
●[1:17:35]. Interpretation of the Intermediate Value Theorem
◉ If a function is continuous on an interval, it cannot "jump" any value between the function values at the
endpoints of the interval.
●[1:19:13]. Application of the Intermediate Value Theorem: Approximating Roots
◉ [1:19:20]. Introduction: The Intermediate Value Theorem can be used to approximate the roots of a
function, that is, the points where the function crosses the x-axis.
◉ [1:19:38]. Procedure:
○ An interval [a, b] is sought where the function changes sign, that is, where 𝒇(a) and 𝒇(b) have
opposite signs.
○ According to the Intermediate Value Theorem, if the function is continuous on [a, b] and 𝒇(a) and 𝒇(b)
have *opposite signs*, then there is at least one root in the interval (a, b).
◉ [1:22:52]. Conclusion: If the signs of 𝒇(a) and 𝒇(b) are different, then there must be at least one root in
the closed interval [a, b].
●[1:23:25]. Example of How to Approximate Roots Using the Intermediate Value Theorem
◉ 𝒇(x) = x^3 - x - 1
◉ [1:23:43]. Procedure:
○ An interval [a, b] is found where the function changes sign. In this example, the interval is used.
○ A table is created with x-values in the interval and the function is evaluated at each x-value.
○ It is observed where the function changes sign. In this example, the sign change occurs between
x = 1.3 and x = 1.4
○ The process is repeated with a smaller interval, in this case [1.3, 1.4], to obtain a more precise
approximation of the root.
◉ [1:26:15]. Conclusion: The process can be repeated with increasingly smaller intervals to obtain an approximation
of the root as precise as desired.
A student sneezes at 18:24 the prof says " Bless you"👌👌🙌
"removable discontinuity" means the whole is removable by changing the function to a piecewise function where the x,y value that fills the hole removes the discontinuity and the that function that has only one value in its domain is c.
Just so the advetisers know: No, I am not tired of searching UA-cam for math help. Yaymath, Kahn academy, and Professor Leonard are better than most things on TV these days.
where is lecture 1.3?
God bless you Prof, you are helping me so much, I am so grateful.
when your own professor recommends you professor leonard yt channel
2 minutes in and I finally I understood a concept that I was stuck on
Thank you so much!! You make the concepts easy to comprehend. Really appreciate it.
now here's my question🗣🗣
you are super amazing prof.
Continuity
●[0:11]. Intuitive Definition of Continuity
◉ A function is continuous if it can be drawn without lifting the pencil from the paper.
◉ It has no holes, jumps, or asymptotes.
●[1:45]. Mathematical Definition of Continuity
◉ A function is continuous at a point C if it meets three conditions:
(ⅰ). The function is defined at that point.
𝒇(c) is defined.
(ⅱ). The limit of the function as x approaches C exists.
Meaning the function approaches the same value from both sides.
(ⅲ). The value of the function at point C is equal to the limit of the function as x approaches C.
lim_(x → c) 𝒇(x) = 𝒇(c)
●[4:41]. Continuity in Terms of Drawing a Function
◉ If you trace a continuous function, you reach a point, fill in the point, and continue without interruptions.
●[5:35]. Graphical Examples of Continuity and Discontinuity
◉[5:37]. Example 1 (ⅰ): A Graph with a Hole at x = C
○ The function is not defined at x = C, so it is not continuous at that point.
○ The limit exists at x = C, but it is not equal to the value of the function (which does not exist).
○ This is an *removeable discontinuity*, as the hole can be "filled in" with a single point.
◉[7:30]. Example 2 (ⅱ): A Graph with a Jump at x = C
○ The function is defined at x = C.
○ The limit does not exist at x = C because the function approaches different values from the left and right.
○ This is a *jump discontinuity*.
◉[9:20]. Example 3 (ⅲ): A Graph with a Vertical Asymptote at x = C
○ The function is or not defined at x = C.
○ The limit exists at x = C and is infinite.
○ This is an *infinite discontinuity*.
◉[11:09]. Summary of Types of Discontinuity
○ There are three main types of discontinuities:
1. Discontinuities with asymptotes.
2. Jump discontinuities.
3. Removeable discontinuities (holes).
○ Removable discontinuities can be "filled in" with a single point.
○ Rational functions typically have removeable discontinuities or asymptotes.
○ Generally, when we encounter a jump discontinuity, the function will be a piecewise function.
●[13:07]Examples, Are these continuous at x = 2?:
◉[13:07]. 𝒇(x) = (x^2 - 4)/(x - 2)
◉[16:20]. 𝓖(x) = {
(x^2 - 4) / (x - 2) if x ≠ 2
3 if x = 2
}
◉[17:36]. 𝒉(x) = {
(x^2 - 4)/(x - 2) if x ≠ 2
4 if x = 2
}
●[18:25]. Continuity on an Interval
◉ If a function is continuous at every point between a and b, it is said to be continuous on the open interval (a, b).
●[19:35]. Continuity at the Endpoints of an Interval
◉ To determine if a function is continuous on a closed interval [a, b], you must verify continuity on the open interval
(a, b) and continuity at the endpoints a and b.
◉ At the endpoints of an interval, one-sided limits are used to verify continuity.
◉ For a function to be continuous at an endpoint, the one-sided limit must exist and be equal to the value of the
function at that endpoint.
●[22:50]. Mathematical Definition of One-Sided Continuity
◉ Continuity from the left: The limit of f(x) as x approaches C from the left must equal f(C).
○ lim_(x → c⁻) 𝒇(x) = 𝒇(c)
◉ Continuity from the right: The limit of f(x) as x approaches C from the right must equal f(C).
○ lim_(x → c⁺) 𝒇(x) = 𝒇(c)
●[25:41]. Function: 𝒇(x) = √(16 - x²) on the Closed Interval [-4, 4].
◉Steps for the Demonstration:
1. Verify continuity on the open interval (-4, 4).
2. Verify continuity from the left at -4.
3. Verify continuity from the right at 4.
◉[29:11]. 1. Demonstration of Continuity on the Open Interval (-4, 4)
○ It must be shown that the limit of f(x) as x approaches C is equal to f(C) for any value of C between -4 and 4.
►lim_(x → c) 𝒇(x) = 𝒇(c)
◉[32:50]. Demonstration of One-Sided Continuity at the Endpoints
◉[34:53]. Conclusion of the Demonstration
● It has been shown that the function is continuous on the open interval and at the endpoints.
● Therefore, the function is continuous on the closed interval [-4, 4].
●[36:00]. Properties of Continuity in Operations with Functions
◉ If f and g are continuous at a point C, then:
○ 𝒇 + 𝓖 is continuous at C.
○ 𝒇 - 𝓖 is continuous at C.
○ 𝒇 * 𝓖 is continuous at C.
◉ 𝒇 / 𝓖 is continuous at C unless 𝓖(C) = 0.
●[38:20]. Discontinuities in Rational Functions
◉ If 𝓖(C) = 0 in a rational function 𝒇 / 𝓖, then there is a discontinuity at C.
◉ The discontinuity can be a hole or an asymptote.
●[40:50]. Continuity of Polynomials
◉ Recall (ⅲ):
○ lim_(x → c) 𝒇(x) = 𝒇(c) ⇔ 𝒇 is continuous at c
◉ Recall the property of polynomial limits: the limit of a polynomial as x
approaches any point can be evaluated simply by substituting x with that point.
○ ∀c ∈ ℝ: lim_(x → c) P(x) = P(c)
◉ This implies that *all polynomials are continuous at all points*.
○ ∀c ∈ ℝ: lim_(x → c) P(x) = P(c) ⇔ P is continuous at every point in ℝ
●[44:10]. Continuity of Rational Functions P(x) / Q(x)
◉ Combining the continuity property of polynomials with the continuity property
of function division, it is concluded that *every rational function is continuous
at all points except where the denominator [Q(x)] is zero*. At that point you have
a *Discontinuity*.
○ [47:15]. Holes and Asymptotes in Rational Functions as Discontinuities.
⑴ Hole (0 / 0): In a rational function, if a common factor can be canceled by factoring the numerator
and the denominator, the discontinuity corresponding to that factor is a hole.
⑵ Asymptote (constant / 0): If a factor of the denominator cannot be canceled, the discontinuity corresponding to
that factor is an asymptote.
◉[50:07]. Example of How to Find Discontinuities in a Rational Function
○ Function: 𝒇(x) = (x² - 4) / (x² - x - 6).
○ Procedure: Discontinuities are found by setting the denominator to zero and solving the equation.
▹ Discontinuities: x = -2 and x = 3.
▹ Factor the function: 𝒇(x) = (x + 2)(x - 2) / (x + 3)(x - 2).
▹ It is observed that the factor (x - 2) cancels out, indicating a hole at x = 2.
▹ The factor (x + 3) does not cancel out, indicating an asymptote at x = 3.
●[55:00]. Demonstration of Continuity of the Absolute Value Function
Procedure: The function is defined piecewise to avoid issues with limits at the endpoints.
◉ Continuity on Open Intervals:
○ For x > 0, the function is simply 𝒇(x) = x, which is a polynomial and therefore continuous.
○ For x < 0, the function is 𝒇(x) = -x, which is also a polynomial and therefore continuous.
◉ Continuity at x = 0:
○ The important thing here is to verify the definition of continuity at x=0; lim_(x → 0) |x| = |0|
○ Continuity from the left: lim_(x → 0⁻) |x| = lim_(x → 0⁻) -x = 0
○ Continuity from the right: lim_(x → 0⁺) |x| = lim_(x → 0⁻) x = 0
I just love Professor's videos and explanations but I can't wrap my head around this. Can anyone explain at 1:04:00 professor made a jump from lim x tends to c f(g(x)) to f(L) and explains that if x is close to c, then g(x) must be equal to L. But, we're supposed to find the limit of f first, not the nested function g. If we're first calculating the limit of g first then f, and then proving that lim x->c f(g(x) = f(Lim x->c g(x)) then isn't the whole proof pointless. Because we're saying that we can put the limit of the composite function inside the nested function but we're literally taking the proof and using it in our steps.
I've noted the same problem. Besides that, at the minute 1:02:25 he says that g(x) is also continuous...
I should have just watched these videos before taking Calculus in college.
I am paying him by watching ads.but still i owe him for such great knowledge.
You are nice.you must be a republican
After watching this I just realized that the limit properties directly correlate to the rules for finding derivatives. Can't believe that I learned everything I needed for Calculus in my first week of class without realizing it.
This lecture is fantastic. I really like your sketches.
I started saying "bless you" after awhile.
1:02:25 - You're assuming that g(x) is also continuous in L? I didn't understand that part.
I really like your teaching style! Are you planning to make some videos about differential equations or linear algebra? Thanks for posting these videos about calculus.
Great...you're are just all I need in mathematics and statistics
Best math class that I ever have omg
omg i understand everything :O
you are the best math teacher
You are amazing! I mean I dotn take Calculus yet, but I still am taking the basics in Algebra 1. Keep making videos! They help a lot.
when the day for VR learning becomes reality this man right here needs to teach the first class.
Everyone is sneezing constantly …but teacher is never 😷 …Those gym hours…😊
Amazing, as always. Thank you so much, Professor!
Thank you Prof. L.
How can I find lecture 1.3? Thank you very much for keep posting your awesome lecture videos.
I was wondering the exact same thing.
@58:30 - Most Students at my Class when the Professor asks something
Good job sir , I like your style... Your teaching motivates me to study calculus
I have an exam in 2 days and I'm playing this in normal speed haha omg
this is me now lol
you truly my hero in math
Nailed it! Thanks prof.
Great video! Very helpful...thank you.
Quality delivery.
ur the only reason for my love in maths
great videos Prof. Leonard
The student of prof Leonard are so frikin lucky!!!
Im a dude just here trying to pass my class but damn his biceps are fkn huge
At 10:00 example 3. Is it that the function dose not have a limit because Infiniti is not a value; it is a term.
Man i needed this.... I hope it works, just downloaded. Thank you
Prof. Leonard, please collaborate with Khan Academy. I will be an asset, I'm sure and that way millions more can benefit from what you offer here. UA-cam is banned in a lot of countries.
P.S why are your students always seem half asleep in your class. I wonder its because they know that they can watch the lectures later when they get home so they don't stay as alert.
Sherlock Holmes Or he could just make his own website and post the videos there too.
+Sherlock Holmes because students are shy.. like me.. i dont wanna speak out unless im sure, and even then im too scared ill be wrong lol
MoonMan moonman, are you enjoying the lunar eclipse right now? It's almost over now anyways.
Sherlock Holmes
i totally forgot about it, i think its over :(