the problem here is we already know that cos cannot get a greater value than 1 so we just do not show the squuze thing maybe I would have done that when I am calm and I have time but not when I am taking an exam. profs just gives zero mark to questions for this kind of mistakes i got 90 marks on calculus 2 and 65 on calculus 1 because this type of stupid things. I am not saying I am not mistaken but taking tests people do this kind of mistakes it is not usefull to give 0 but when I do this on the homework yeah give me zero. The most annoying part is profs. do the worst when they giveaway their notes and it takes hours to understand where this "fi" or that "alpha" came from like they don't actually wants us to understand the topic.
In Germany the squeeze theorem is sometimes also called "Der Satz von den zwei Polizisten" which translates to "The Theorem of the two policemen". In my eyes a way cooler name.
Correct me if I'm wrong, but I think that the function sin(1/x) and xsin(1/x) if graph will have the same number of solutions in a particular domain. This means that we can multiply x to both sides and be sure that we are not introducing any new solutions for x
0:51 this is the same question I had got in my test yesterday, although it asked to check the continuity of f and f' at x=0, and differentiability too.😮
I did go into high school calculus already self taught in a fair amount of college level calculus. Going through the basics again was actually fun. I never was a "just memorize the rules" kind of person. I loved understanding and playing out the mechanics behind the rules. Totally worth looking at it from the ground up all over again. And I still do it now for fun. It's exciting going through deriving things.
I'm glad he taught us "Sequeeze" here because I would have done the same mindless solution he was mocking about by taking the derivative and then scratching my head and giving up vs realizing "Sequeeze" Calculus evaluating as limit x ---> 0 in this, I thought was a piece-wise continuous function but indeterminate at x = 0. I learned something additional today when, I the student, also was saying, "I know this! Wait? No I don't!" 😂
I’m curious, why is the squeeze theorem necessary to prove that 2xsin(1/x) = 0 for x = 0? Isn’t it obvious that no matter what x is sin is bounded between -1 and 1 and 0 multiplied by anything between -1 and 1 is zero?
@@BossX2243... The derivative chain rule left - cos(1/x) term he mentions as the limit of the derivative is -1 to 1. We don't know out to 1/x part going to infinity where Cos (1/x) term ends at. ... So because he teaches the proof of the derivative is better in determining what f(x) does near x=0. It was a good lecture! 👍
When you got to the Squeeze Theorem, you were indeed using the main part of the definition of the derivative, but you kind of glossed over the other part, showing that it is in fact differentiable. (Also, when you multiplied by x, you were assuming that x > 0, because you said that -1x < xf < 1x.)
@@davidebic i think he referred to the part where f'(x) exists when the limits to the right and left of f'(x) also exist and they are equal with f'(x); f'(x) by itself doesn't mean anything - he should have started with the definition to show that f'(0) = 0, follow up with lim x->0 f'(x) doesn't exist, and conclude that f'(0) doesn't exist either
@@radadadadee The limit of x*sin(1/x^n) for any n as x approaches 0 is indeed 0 though. And it can be shown through squeeze theorem. I don't get what you mean.
@@radadadadee sin is a limited function that assumes values within [-1, 1] and x is a function that tends to 0 as x approaches 0, so it indeed is 0 for any 1/x^n. It's actually 0 for ANY function.
You are a master in Calculus. I understand everything you teach us and with funny humor. I'm still learning advanced Calculus new thoughts watching many of your videos! 👍
Ah 15+ years of using this def and knowing how derivatives are made, and I never thought of deriving that by hand... I'm feeling enlightened and humbled all at once 😮
fun fact In many Countries the Squeeze/Sandwich Theorem is also called the Gendarme Theorem (in case you don't know, the Gendarmes it's a military army who has the same duties as the police) And the metaphor is very intuitive: the outer functions (in this case -x and x) are the two gendarmes who are going to catch the criminal represented by the inner function(in this case xsin(1/x))
Yeah definition of derivative is underrated. It’s not even in our syllabus and is just briefly talked about as “for understanding”. But once you tell students that, 99% of them will just immediately tune out unfortunately. Another good example where the standard differentiation techniques fail is when we try to find the derivative of a function to a function power when the graph is at 0. The standard method of logarithmic differentiation wont work because we will end up with ln0 somewhere, but the definition of derivative can get us to the answer
@@farfa2937 Did you completely miss the point I was trying to get at? The point was that understanding and knowing how to use the definition of derivative is just as important as the so called shortcut methods . Especially when you want to prove theorems involving derivatives and differentiability, the definition is the only way to go.
Believe it or not, this kind of analysis is very useful for my engineering class and I'm glad this video got recommended to me. If y'all like studying audio, control systems, or signals in general look up "Final value theorem" or "Initial value theorem." That sin(1/x) function is really interesting and I'm not sure if it would even be possible to find the final value for that. But I like how you showed the analysis of each limit and using the squeeze theorem to evaluate the results of a completely different function which it's actually possible to find the limit of. Keep up these amazing explanations for struggling calculus and struggling engineers who forgot their calculus and need to dust off the cobwebs 😅🤣!
My Calc 1 teacher insisted of demonstrated all the common derivative results. My head couldn't really keep up but I could tell he cared about doing it the correct way
my teacher always puts a “by definition” exercise in tests, i really like it because those are easy if you pay attention in class cuz she usually demonstrates these things and makes us think about each step before she writes it down in the blackboard
I think the Sandwich or Squeeze Theorem isn’t absolutely necessary to solve this limit: as x goes to 0, 1/x tends to ∞; however since the range of sin (x) is [-1, 1], as 1/x approaches ∞, sin (1/x) will take some finite value. But with x going to zero, the required limit is clearly of the form ZERO APPROACHING TIMES SOMETHING FINITE FROM -1 TO 1 which isn’t even an indeterminate form. Hence, the limiting value is zero.
You single handed Help me to Pass my Advanced Mathematics Thank you so Much I am so happy to see you with 1 million subscribers i really enjoyed your videos on 100 integrals back in 2019
The inequality stays true as long as you use the mechanics of inequalities and properly flip the directions when multiplying by a negative number, ie. x>5 multiplying through by -1 gives -x
@@command0_ I understand, but I am positing that because you used x=-6/pi which is a negative x value then your inequalities should flip so the proper statement is: 6/pi >= 3/pi >= -6/pi which is true.
Thank you. I have a lot to review for when I return to continue my bachelor's in math. I'm currently studying for the CCNA networking exam and I haven't been in school since right before the pandemic. I'm doing the CCNA so I can land a decent paying job to pay for school which probably isn't the best way to go about it but for some reason I feel I have to do it this way.
I know you have thousands of requests for every video you do, but could you please go over some moderately difficult related rates / optimization problems? I have the AP Calculus BC test in a couple months and that's one of the sections that I'm least confident in. Also - great video & explanations!
@@SuperMaDBrothers I used to think like you when I was younger. I used to think that I barely need any practice if I already understand the math. Indeed it used to be like that back in my middle school and high school days. Once you get to higher math at a university (if you do choose to be a math major), you will meet proofs classes like real analysis (basically proving every single theorem in calculus), abstract algebra, topology, and more (and even something seemingly benign like graph theory would require you to prove some theorems)... then you will realize how important practice could be. Even if you understand every theorem, you can easily forget all the proofs if you don't practice proving them. Real analysis homework and exams are about proving, not just reciting and using theorems. Even if you somehow managed to master proving every theorem in the book, you are yet to master the proof exercises at the end of every chapter, where you have to use the theorems to prove other things. The one in this video is a relatively easy example of a real analysis problem. (Think about it, this video uses the squeeze theorem, but can you prove the squeeze theorem? Yes it's intuitive, it makes sense, it seems like it doesn't need proving, but it does need to be proved in real analysis, using the definition of the limit, etc) Not to mention, university/college courses/classes tend to be quite fast-paced...
4:55. While 0 × DNE isn't 0, sin (1/x) oscillates between 1 and -1. There won't be a 0 × infinity case. So the limit is 0. In general: limx->0 f(x)sin(g(x)) = 0 if limx->0 f(x) = 0, even if limx->0 g(x) is infinity or -infinity
In Italy we call it the "Carabinieri theorem": police theorem for not italian speakers. That is because the side functions are like 2 policemen forcing a person to go through (the central function)
This also explains why one should take care when dealing with C^1 functions. The function f(x) = x^2 sin(1/x) with f(0) = 0 is continuous and differentiable for every real x but the derivative f' is not continuous, so f does not belong to C^1.
The squeeze theorem doesn't state "if g(x)a h(x) = L, and limx->a g(x)a f(x)a h(x), then limx->a f(x) = L." In fact any x sandwiched between two same numbers L, by simple logic will be equal to L. Instead, the theorem states"if g(x)a h(x) = L, then limx->a f(x) = L." So your way of taking the limit of all 3 sides of the inequality is wrong, unless if you say you can take the limit of all 3 sides BECAUSE of the squeeze theorem, in which is true, as implied by its statement
At 5:00 when you say that you CANT conclude what something simplifies to if you have 0•DNE, how come you concluded that f(0) up above was 0 when it has the same circumstances as the new one? Im assuming it's because the f(0) has an x^2 which in the way I imagine it, approaches 0 quicker than whatever else is being multiplied approaches infinity.. or something else. Im not sure and seriously need an explanation!
I always thought that (sinx)/x is zero as limit of x approaches 0 from the sandwich theorem, but here it's interesting that x sin(1/x) as x tends to 0 which can also be written as lim x->0 [ sin(1/x) ]/(1/x) is equal to one right, how is it going to be zero? Edit : Oh gotchu!!! So that limit is equal to one, but they are asking for the derivative and it becomes zero right? Am i right?
if we let t = 1/x, then lim {x-> 0} [ sin(1/x) ] / (1/x) = lim {t -> inf} sin(t) / t We know lim {t -> 0} sin(t) / t = 1 but this is NOT what we're looking at here. The limit is taken as t goes up to infinity, rather than down to zero, so this result doesn't apply.
1/x is not approaching zero when x approaching zero. we can only write lim x ->0 sin(f(x))/f(x)=1 only if the function f(x) approaches zero when x approaches zero
Hi, a few months ago I "designed" (found) a nice function h(x) similar to this function here, where each derivative at x = 0 is 0: d^n/dx^n h(0) = 0 for all n in IN. Therefore, no one can decide whether the function has an extremum or inflection point at x = 0 xD The definition was: h: IR -> IR, x -> h(x) = exp(-1/x^2) * sin(1/x) for x 0 and h(x) = 0 for x = 0. Maybe you like this property of a function too ;)
Well, since you can find two sequences xn and yn that tend towards 0, h(xn) is strictly positive and h(yn) is strictly negative, it isn't an extremum. For example, it could be xn=1/(2kπ+π/2) and yn=-xn. Since convexity is equivalent to the monotony of the derivative (given that it exists), I imagine a similar argument would be possible for the derivative (with the same result according to Desmos), but I'm not going to attempt it right now, as I must really go to sleep.
I think it would have been beneficial to explain that the reason the product rule doesn't work is because it assumes that the component functions x^2 and sin(1/x) are both differentiable at x=0, which is untrue for the latter (more specifically, there is an essential singularity there). Had to get chat GPT to explain that one.
I've just barely entered high school and have no idea what any of the terms shown in the video even remotely mean, yet it is quite interesting to observe it either way.
What I'd really like to see is a proof that f'(x) = 2xsin(1/x) - cos(1/x) using the definition of derivative.... I got the 2xsin(1/x), but somehow managed to lose the -cos(1/x). What did I do wrong??
product rule is [u(x)v(x)]' = u'(x)v(x) + v'(x)u(x) You differentiated x^2 just fine but there's another term in the answer where you differentiate sin(1/x) f(x) = x^2 sin(1/x) f'(x) = [x^2]' sin(1/x) + x^2 [sin(1/x)]' = 2x sin(1/x) + x^2 (cos(1/x)) (-1/x^2) = 2x sin(1/x) - cos(1/x)
When you multiply both sides of the inequality by x, you have to reverse the direction of the inequalities whenever x is negative. Otherwise, suppose for example that x=-3. Then your equation says 3
Firstly I didn't realize what is "squeeze theorem", because in our country we call it literally "theorem about two guards/escorts". And I think that the second variant is easier understood. Also thanks for the video.
there is a mistake, you cannot multiply an innequality by X. you must take the limit from the left of 0 (positive number inequality is preserved) and from the left (negative number, inequality signs are changed) and a the end you obtain the desired result.
5:44 is it not possible to just multiply by x since we don't know if it's positive or negative? Since it's the function is defined on all R exept 0 we can't tell if it's positive or negative
0:26 it’s because we want epsilon delta 😜 Now that I think about it, is it possible for smth to be impossible to answer with derivative rules or the definition of the derivative but is possible with the epsilon delta definition?
Here's a interesting question that I've been wondering even since I learnt trig and Heron's formula: With cosine rule to find an angle and sine function triangle formula we can find the area of a triangle with its 3 sides. With heron's formula we can also find the area of a triangle with its 3 sides. Thus, if we equate these two do we: a. Get an identity? b. Get the ratio between 3 sides? (Which I know, there is none) I've put it in Wolfram Alpha, it took me like 5 minutes but there was no meaningful answer.
Nice observation. I remember wondering about this myself a sone time ago, and having done some algebra to check that the two formulae are equivalent. So here we go. Suppose we have a triangle with sides a, b, c, and that its area is A. Heron's formula states that A=√[s(s-a)(s-b)(s-c)] where s=(a+b+c)/2. On the other hand, we can calculate cos C in terms of a, b, c using the cosine rule, calculate sin C from cos C using the Pythagorean identity cos²C+sin²C=1 and the fact that sin C≥0, and finally calculate the area using the formula A=½ab sin C. Our method is to convert the trignometric formula into a formula just involving the sides, using the cosine rule and the Pythagorean identity, and show we end up with Heron's formula. The argument applies whether we are given the three sides a, b, c or two sides a, b and angle C. By the cosine rule we have c²=a²+b²-2ab cos C 2ab cos C=a²+b²-c² We now square both sides so as to be able to use the Pythagorean identity in the form cos²C=1-sin²C: (2ab cos C)²=(a²+b²-c²)² 4a²b²cos²C=(a²+b²-c²)² By the Pythagorean identity 4a²b²cos²C=4a²b²(1-sin²C)=4a²b²-4a²b²sin²C So 4a²b²sin²C=4a²b²-4a²b²cos²C Applying our equation above for 4a²b²cos²C we get 4a²b²sin²C=4a²b²-(a²+b²-c²)² =(2ab)²-(a²+b²-c²)² =[2ab+(a²+b²-c²)][2ab-(a²+b²-c²)] =[a²+2ab+b²-c²][2ab-a²-b²+c²] =[(a²+2ab+b²)-c²][c²-(a²-2ab+b²)] =[(a+b)²-c²][c²-(a-b)²] =(a+b+c)(a+b-c)[c+(a-b)][c-(a-b)] =(a+b+c)(a+b-c)(c+a-b)(c+b-a) =(a+b+c)(a+b+c-2c)(c+a+b-2b)(c+b+a-2a) =2s(2s-2c)(2s-2b)(2s-2a) 4a²b²sin²C=16s(s-c)(s-b)(s-a) Dividing both sides by 16: ¼a²b²sin²C=s(s-a)(s-b)(s-c) (½ab sin C)²=s(s-a)(s-b)(s-c) Taking the non-negative square root of each side (noting sin C≥0) ½ab sin C=√[s(s-a)(s-b)(s-c)] which shows the equivalence of the two formulae for the area of a triangle.
BTW, you can get a nice identity from this - a way to calculate sin A directly from the three side lengths: As ½bc sin A=√[s(s-a)(s-b)(s-c)], we get sin A=2√[s(s-a)(s-b)(s-c)]/bc For example, for an equilateral triangle of side 2, we get s=3, s-a=s-b=s-c=1, so sin A=2√[3(s-a)(s-b)(s-c)]/bc =2√[3×1×1×1]/2×2 =√3/2, as expected.
Our whole class was definitely taught the derivative of sin before we were taught the proof behind it, anyway I found it’s better to just analyze it like how the lim x->0 of sin (x)/x=1 then you know the derivative at 0 is 1 and you can find the max at pi/2 so you know the derivative there is 0, from there it’s pretty easy for anyone to see the derivative is cos
Is it strictly necessary to define f(0)=0, or can we just omit it by declaring f(x) is continuous? Clearly the squeeze theorem (sequeeze?) tells us the limit of f(x) as x->0 is 0, so defining x(0)=0 can just be filled in by continuous extension.
Since f is sandwiched between x² and -x², both of which have zero value and derivative at x=0, you can conclude f'(0) = 0 by squeeze theorem (applied to the definition of the derivative on all three functions). P.S. Ok that is almost identical to the video, just without cancelling x in the denominator.
Could this be a way to conclude that f'(0)=0, and if so, does this method have a name? kind of a squeeze theorem but for derivatives, maybe? f(x) is bounded on top by g(x)=x^2 and on the bottom by h(x)=-x^2 Given that both g'(x) and h'(x) are continuous and they are equal at x=0, then that must be the derivative of f(x) at that point.
1) ask those students: why did you come here if you already know it? 2) call those students parrots that spit textbook back at the examiner but have no clue where and how they are going to use it. Either method works :)
1) "I have to be here" 2) there are two variations: "this is untrue, I understand perfectly well where to use it" or "so what?" Also insulting your students isn't exactly going to motivate them
could you explain why in second Order diff equations with complex roots there Is no i in the solution?Is It included in the constant C?shouldnt It be only real?
Only if it satisfies the previous step. And it is exactly what the squeeze theorem says. In this example looking at the squeeze theorem around 0, Starting with some interval I, where 0 ∈ I, if you happen to know that a
I’m curious, why is the squeeze theorem necessary to prove that 2xsin(1/x) = 0 for x = 0? Isn’t it obvious that no matter what x is sin is bounded between -1 and 1 and 0 multiplied by anything between -1 and 1 is zero?
Yes, the squeeze theorem is just a formal way to show it, but your intuition is right and many mathematicians would also just say it equals 0 and not bat an eye. But you don't need squeeze theorem: Theorem: If lim x->0 f(x)=0 and g(x) is bounded, then lim x->0 f(x)g(x)=0. Proof: Since g is bounded, let M be such that |g| < M. Let e>0. Since lim x->0 f(x)=0, we can find D such that when |x|
“something between -1 and +1” is of course not known but is a limited number so when is multiplied by 0 is not undefined, it MUST BE 0 … the squeeze theorem could not give a different value 🤷🏻♂️
I have 2 questions. Doesn't this assume that f(x) = x^2 * sin(1/x) as x approaches 0, an assumption with no reasoning? Also, is the first method of checking the derivative from the left and right flawed, or does is this method inconclusive if it says the limit DNE?
The general theorem says: if you have a continuous function in an interval, such that the derivative always exists expect at most in a point x0. IF the limits of the derivative from left and right both exists and are equal, then the derivative exists also I n x0 and coincides with those limits. Here there is the big assumption that the function is continoisv not only in x0, but in an interval surrounding it, and that the only non-differentiable point is at most only x0 Essantially it applies to continous piece-wise functions
guys i would like to say that the function is the thumbnail is oscillating therefore discontinuous and in-differentiable. so f’(0) = undefined. LESSSSS GO SON
My Calculus I professor called it the "pinching theorem." He was from India and his accent was difficult for me to understand at times. I thought for a while he was saying "peachy theorem."
Learn more calculus on Brilliant: brilliant.org/blackpenredpen/ (now with a 30-day free trial)
But now, I can say "I know it already" to this equation!
the problem here is we already know that cos cannot get a greater value than 1 so we just do not show the squuze thing maybe I would have done that when I am calm and I have time but not when I am taking an exam. profs just gives zero mark to questions for this kind of mistakes i got 90 marks on calculus 2 and 65 on calculus 1 because this type of stupid things. I am not saying I am not mistaken but taking tests people do this kind of mistakes it is not usefull to give 0 but when I do this on the homework yeah give me zero. The most annoying part is profs. do the worst when they giveaway their notes and it takes hours to understand where this "fi" or that "alpha" came from like they don't actually wants us to understand the topic.
Can i send you a problem
Its a very complicated logarithme equation
Replay me as soon as possible
My calc teacher caught the class out with this exact example last week.
Great minds, they say
this was on my ontario grade 12 advanced functions test lol
He has a whole shelf of markers 😂
In Germany the squeeze theorem is sometimes also called "Der Satz von den zwei Polizisten" which translates to "The Theorem of the two policemen". In my eyes a way cooler name.
Théorème des gendarmes, same idea... But I prefer sandwiches, as they are edible 🙂
In Finnish it's called The Strangulation Principle... or "kuristusperiaate" lol
Wouldn’t work in America 💀
Same thing in Russia
same in italy :)
5:50 Multiplying by x is technically incorrect. It works only for x>0. You should seperatly check what happens for x
Or you could write it as -|x| < x sin(1/x) < |x|
Yes, man! You spot the loophole! 😘
Multiplication by negative values of x yields the same inequalities in absolute value.
Correct me if I'm wrong, but I think that the function sin(1/x) and xsin(1/x) if graph will have the same number of solutions in a particular domain. This means that we can multiply x to both sides and be sure that we are not introducing any new solutions for x
@@rasheedmohammed2227 Can you explain how you deduced that they have the same amount of solutions? Isn't there ambiguity when x is around zero?
I love the Squeeze Theorem.
There is something hugely satisfying about catching a value between two other values like this.
@@radadadadee It does go to zero for any real power of p
its also called the sandwich theorem !!
Sequeeze theorem aka sayndwich theorem lol.
0:51 this is the same question I had got in my test yesterday, although it asked to check the continuity of f and f' at x=0, and differentiability too.😮
e same
youtubeeem.com/MTkb8MBftFk
you. I have a lot to review for when I return to continue my bachelor's
me too in france lol, it's a great exemple
I did go into high school calculus already self taught in a fair amount of college level calculus. Going through the basics again was actually fun. I never was a "just memorize the rules" kind of person. I loved understanding and playing out the mechanics behind the rules. Totally worth looking at it from the ground up all over again. And I still do it now for fun. It's exciting going through deriving things.
Am I the only one seeing "sequeeze" instead of "squeeze" or is it a ploy to get me to comment 😅
😅 You're not alone. I was just about to say the same thing myself.
Same here
I'm glad he taught us "Sequeeze" here because I would have done the same mindless solution he was mocking about by taking the derivative and then scratching my head and giving up vs realizing "Sequeeze" Calculus evaluating as limit x ---> 0 in this, I thought was a piece-wise continuous function but indeterminate at x = 0. I learned something additional today when, I the student, also was saying, "I know this! Wait? No I don't!" 😂
I’m curious, why is the squeeze theorem necessary to prove that 2xsin(1/x) = 0 for x = 0? Isn’t it obvious that no matter what x is sin is bounded between -1 and 1 and 0 multiplied by anything between -1 and 1 is zero?
@@BossX2243... The derivative chain rule left - cos(1/x) term he mentions as the limit of the derivative is -1 to 1. We don't know out to 1/x part going to infinity where Cos (1/x) term ends at. ... So because he teaches the proof of the derivative is better in determining what f(x) does near x=0. It was a good lecture! 👍
When you got to the Squeeze Theorem, you were indeed using the main part of the definition of the derivative, but you kind of glossed over the other part, showing that it is in fact differentiable. (Also, when you multiplied by x, you were assuming that x > 0, because you said that -1x < xf < 1x.)
In 1-D differentiable is the same as derivable.
@@davidebic i think he referred to the part where f'(x) exists when the limits to the right and left of f'(x) also exist and they are equal with f'(x); f'(x) by itself doesn't mean anything - he should have started with the definition to show that f'(0) = 0, follow up with lim x->0 f'(x) doesn't exist, and conclude that f'(0) doesn't exist either
Isn't f differentiable in 0 if that limit exists? Since he found f'(0) = 0, f is differentiable in 0? I don't get what he misses there.
@@radadadadee The limit of x*sin(1/x^n) for any n as x approaches 0 is indeed 0 though. And it can be shown through squeeze theorem. I don't get what you mean.
@@radadadadee sin is a limited function that assumes values within [-1, 1] and x is a function that tends to 0 as x approaches 0, so it indeed is 0 for any 1/x^n. It's actually 0 for ANY function.
You are a master in Calculus. I understand everything you teach us and with funny humor. I'm still learning advanced Calculus new thoughts watching many of your videos! 👍
現在太顯瘦了....當年這劇本實在好到不行
youtubeeem.com/DEuJI6xmOdO
亂演!!就憑共產黨!不用理由好人都可以靠起來消失!! 真以為中共有法治喔,那是人治
I could see myself taking some of these shortcuts, so thank you for showing me how to do it the right way!
Wow. This was honestly an awesome problem on the utility and value of the limit definition of the derivative! So cool 😊
Thanks, Sergio!
Ah 15+ years of using this def and knowing how derivatives are made, and I never thought of deriving that by hand... I'm feeling enlightened and humbled all at once 😮
Great video on a neat problem, but I was also mesmerized by your marker-switching technique. Absolutely stellar.
I've taken masters level math courses and I'm constantly forgetting the fundamentals. This isn't a burden but a gift. Thank you
a friend of every analysis student. this function also illustrates that the derivative of a function need not be continuous
fun fact
In many Countries the Squeeze/Sandwich Theorem is also called the Gendarme Theorem (in case you don't know, the Gendarmes it's a military army who has the same duties as the police)
And the metaphor is very intuitive: the outer functions (in this case -x and x) are the two gendarmes who are going to catch the criminal represented by the inner function(in this case xsin(1/x))
Drunk man and two policemen theorem.
Yeah definition of derivative is underrated. It’s not even in our syllabus and is just briefly talked about as “for understanding”. But once you tell students that, 99% of them will just immediately tune out unfortunately.
Another good example where the standard differentiation techniques fail is when we try to find the derivative of a function to a function power when the graph is at 0. The standard method of logarithmic differentiation wont work because we will end up with ln0 somewhere, but the definition of derivative can get us to the answer
Well, it's what you would call a waste of time. The functions that fail the easy rules are not organic, they were just built to fail - like this one.
@@farfa2937 Did you completely miss the point I was trying to get at? The point was that understanding and knowing how to use the definition of derivative is just as important as the so called shortcut methods .
Especially when you want to prove theorems involving derivatives and differentiability, the definition is the only way to go.
@@farfa2937Incorrect. These types of functions show up in quantum mechanics (seldomly but they do show up).
TFW someone claims all non-elementary functions are unnatural.
Believe it or not, this kind of analysis is very useful for my engineering class and I'm glad this video got recommended to me. If y'all like studying audio, control systems, or signals in general look up "Final value theorem" or "Initial value theorem." That sin(1/x) function is really interesting and I'm not sure if it would even be possible to find the final value for that. But I like how you showed the analysis of each limit and using the squeeze theorem to evaluate the results of a completely different function which it's actually possible to find the limit of. Keep up these amazing explanations for struggling calculus and struggling engineers who forgot their calculus and need to dust off the cobwebs 😅🤣!
My Calc 1 teacher insisted of demonstrated all the common derivative results. My head couldn't really keep up but I could tell he cared about doing it the correct way
Sin( 1/x ) may be undefined but it is bounded. In x * sin( 1/x ) as x->0 we get 0* a number = 0.
my teacher always puts a “by definition” exercise in tests, i really like it because those are easy if you pay attention in class cuz she usually demonstrates these things and makes us think about each step before she writes it down in the blackboard
I think the Sandwich or Squeeze Theorem isn’t absolutely necessary to solve this limit: as x goes to 0, 1/x tends to ∞; however since the range of sin (x) is [-1, 1], as 1/x approaches ∞, sin (1/x) will take some finite value. But with x going to zero, the required limit is clearly of the form ZERO APPROACHING TIMES SOMETHING FINITE FROM -1 TO 1 which isn’t even an indeterminate form. Hence, the limiting value is zero.
Yeah I also made a comment about that mentioning it.
Glad to have reviewed this, completely forgot about it
You single handed Help me to Pass my Advanced Mathematics Thank you so Much I am so happy to see you with 1 million subscribers i really enjoyed your videos on 100 integrals back in 2019
Just a small correction: The inequality at 5:50 does not hold for x
The inequality stays true as long as you use the mechanics of inequalities and properly flip the directions when multiplying by a negative number, ie. x>5 multiplying through by -1 gives -x
@@breakfastbat I agree with you that inequalities work in this way when multiplying by -1, however I am simply saying that the inequality -x
Yeah I was thinking about this too
@@command0_ I understand, but I am positing that because you used x=-6/pi which is a negative x value then your inequalities should flip so the proper statement is: 6/pi >= 3/pi >= -6/pi which is true.
The generalized solution to which is either your absolute value solution or defining one inequality for x>=0 and another with the signs flipped for x
Thank you. I have a lot to review for when I return to continue my bachelor's in math. I'm currently studying for the CCNA networking exam and I haven't been in school since right before the pandemic. I'm doing the CCNA so I can land a decent paying job to pay for school which probably isn't the best way to go about it but for some reason I feel I have to do it this way.
That's a great problem to do along side the sincx function and then introduce students to information processing theory and spectroscopy.
I know you have thousands of requests for every video you do, but could you please go over some moderately difficult related rates / optimization problems? I have the AP Calculus BC test in a couple months and that's one of the sections that I'm least confident in. Also - great video & explanations!
recognize me?
Also, if you get the math, you shouldn't need to practice :P that's just dumb.
@@SuperMaDBrothers
I used to think like you when I was younger. I used to think that I barely need any practice if I already understand the math. Indeed it used to be like that back in my middle school and high school days. Once you get to higher math at a university (if you do choose to be a math major), you will meet proofs classes like real analysis (basically proving every single theorem in calculus), abstract algebra, topology, and more (and even something seemingly benign like graph theory would require you to prove some theorems)... then you will realize how important practice could be.
Even if you understand every theorem, you can easily forget all the proofs if you don't practice proving them. Real analysis homework and exams are about proving, not just reciting and using theorems. Even if you somehow managed to master proving every theorem in the book, you are yet to master the proof exercises at the end of every chapter, where you have to use the theorems to prove other things. The one in this video is a relatively easy example of a real analysis problem. (Think about it, this video uses the squeeze theorem, but can you prove the squeeze theorem? Yes it's intuitive, it makes sense, it seems like it doesn't need proving, but it does need to be proved in real analysis, using the definition of the limit, etc)
Not to mention, university/college courses/classes tend to be quite fast-paced...
@@bismajoyosumarto1237 i graduated from an ivy league and work in quantum computing
@@SuperMaDBrothers
Oh nice, were you a math major or?
@@bismajoyosumarto1237 nah, physics. Pure math is dumb lol it’s just a tool, physics realizes math is a tool and takes everything way further
if you consider x on the interval -1 < x < 1 , ... and that -1
4:55. While 0 × DNE isn't 0, sin (1/x) oscillates between 1 and -1. There won't be a 0 × infinity case. So the limit is 0. In general: limx->0 f(x)sin(g(x)) = 0 if limx->0 f(x) = 0, even if limx->0 g(x) is infinity or -infinity
In Italy we call it the "Carabinieri theorem": police theorem for not italian speakers. That is because the side functions are like 2 policemen forcing a person to go through (the central function)
I'm currently starting Calculus 1, and I have to force myself to stay quiet about some more advanced stuff I learned from channels like you XD
😆
This also explains why one should take care when dealing with C^1 functions.
The function f(x) = x^2 sin(1/x) with f(0) = 0 is continuous and differentiable for every real x but the derivative f' is not continuous, so f does not belong to C^1.
I think in 05:40, it should be -|x|
you know he a calc teacher bc of all those Expo markers he got in the back
The squeeze theorem doesn't state "if g(x)a h(x) = L, and limx->a g(x)a f(x)a h(x), then limx->a f(x) = L." In fact any x sandwiched between two same numbers L, by simple logic will be equal to L. Instead, the theorem states"if g(x)a h(x) = L, then limx->a f(x) = L." So your way of taking the limit of all 3 sides of the inequality is wrong, unless if you say you can take the limit of all 3 sides BECAUSE of the squeeze theorem, in which is true, as implied by its statement
5:34 you forgot about negative x
At 5:00 when you say that you CANT conclude what something simplifies to if you have 0•DNE, how come you concluded that f(0) up above was 0 when it has the same circumstances as the new one?
Im assuming it's because the f(0) has an x^2 which in the way I imagine it, approaches 0 quicker than whatever else is being multiplied approaches infinity.. or something else. Im not sure and seriously need an explanation!
f(0) was defined to be 0. That’s why f(0)=0.
@@blackpenredpen Oooooh haha, thanks for clarifying! Keep doing what you do!
"...is by forcing them"
Well, that's how I learned most of what I know now.
The squeeze theorem is called as sandwich theorem in India
I always thought that (sinx)/x is zero as limit of x approaches 0 from the sandwich theorem, but here it's interesting that x sin(1/x) as x tends to 0 which can also be written as lim x->0 [ sin(1/x) ]/(1/x) is equal to one right, how is it going to be zero?
Edit : Oh gotchu!!! So that limit is equal to one, but they are asking for the derivative and it becomes zero right? Am i right?
Why don't you edit the original comment an repost it, you are confusing me if i should consider your main comment or the edit.
if we let t = 1/x, then
lim {x-> 0} [ sin(1/x) ] / (1/x) = lim {t -> inf} sin(t) / t
We know
lim {t -> 0} sin(t) / t = 1
but this is NOT what we're looking at here. The limit is taken as t goes up to infinity, rather than down to zero, so this result doesn't apply.
1/x is not approaching zero when x approaching zero. we can only write lim x ->0 sin(f(x))/f(x)=1 only if the function f(x) approaches zero when x approaches zero
Is it sequeeze theorem? I noticed you spelled it that way twice. It sounds like you are saying squeeze theorem, but I’m not sure. Thanks
Hi, a few months ago I "designed" (found) a nice function h(x) similar to this function here, where each derivative at x = 0 is 0: d^n/dx^n h(0) = 0 for all n in IN. Therefore, no one can decide whether the function has an extremum or inflection point at x = 0 xD The definition was: h: IR -> IR, x -> h(x) = exp(-1/x^2) * sin(1/x) for x 0 and h(x) = 0 for x = 0. Maybe you like this property of a function too ;)
Well, since you can find two sequences xn and yn that tend towards 0, h(xn) is strictly positive and h(yn) is strictly negative, it isn't an extremum. For example, it could be xn=1/(2kπ+π/2) and yn=-xn. Since convexity is equivalent to the monotony of the derivative (given that it exists), I imagine a similar argument would be possible for the derivative (with the same result according to Desmos), but I'm not going to attempt it right now, as I must really go to sleep.
Oh, a nice example of a function that is smooth but not continuously smooth.
In general if you have f(x)
I live for this man saying "this and that", nothing more satisfying
I think it would have been beneficial to explain that the reason the product rule doesn't work is because it assumes that the component functions x^2 and sin(1/x) are both differentiable at x=0, which is untrue for the latter (more specifically, there is an essential singularity there). Had to get chat GPT to explain that one.
No, primarily the product rule doesn't work because f is not defined as the product of two functions - the value at 0 is set manually.
I've just barely entered high school and have no idea what any of the terms shown in the video even remotely mean, yet it is quite interesting to observe it either way.
x²sin(1/x) = 0 with x≠0 means sin(1/x) = 0, so 1/x = arcsin(0) = nπ, so x = 1/(nπ). With n, an interger.
What I'd really like to see is a proof that f'(x) = 2xsin(1/x) - cos(1/x) using the definition of derivative.... I got the 2xsin(1/x), but somehow managed to lose the -cos(1/x). What did I do wrong??
product rule is [u(x)v(x)]' = u'(x)v(x) + v'(x)u(x)
You differentiated x^2 just fine but there's another term in the answer where you differentiate sin(1/x)
f(x) = x^2 sin(1/x)
f'(x) = [x^2]' sin(1/x) + x^2 [sin(1/x)]'
= 2x sin(1/x) + x^2 (cos(1/x)) (-1/x^2)
= 2x sin(1/x) - cos(1/x)
@@inazumapi2428 I'm talking about a proof by using definition of derivative.
When you multiply both sides of the inequality by x, you have to reverse the direction of the inequalities whenever x is negative. Otherwise, suppose for example that x=-3. Then your equation says 3
0:44 That's the first example our teacher used when he explained the importance of the derivative definition lol
Firstly I didn't realize what is "squeeze theorem", because in our country we call it literally "theorem about two guards/escorts". And I think that the second variant is easier understood.
Also thanks for the video.
The interesting thing about this function is Taylor expansion for it which does not apply due to the finite remaining term
Thank you for the video. I have a question. Can you prove that 2^x>2^x, for x>4, using derivatives.
I would also appreciate if you solved that!
Proving that would be quite disturbing.
there is a mistake, you cannot multiply an innequality by X. you must take the limit from the left of 0 (positive number inequality is preserved) and from the left (negative number, inequality signs are changed) and a the end you obtain the desired result.
I like your teaching way your teaching way is different from all other I easily understand it ❤thank you teacher 😊
5:44 is it not possible to just multiply by x since we don't know if it's positive or negative? Since it's the function is defined on all R exept 0 we can't tell if it's positive or negative
0:26 it’s because we want epsilon delta 😜
Now that I think about it, is it possible for smth to be impossible to answer with derivative rules or the definition of the derivative but is possible with the epsilon delta definition?
Here's a interesting question that I've been wondering even since I learnt trig and Heron's formula:
With cosine rule to find an angle and sine function triangle formula we can find the area of a triangle with its 3 sides.
With heron's formula we can also find the area of a triangle with its 3 sides.
Thus, if we equate these two do we: a. Get an identity? b. Get the ratio between 3 sides? (Which I know, there is none)
I've put it in Wolfram Alpha, it took me like 5 minutes but there was no meaningful answer.
Nice observation.
I remember wondering about this myself a sone time ago, and having done some algebra to check that the two formulae are equivalent.
So here we go.
Suppose we have a triangle with sides a, b, c, and that its area is A.
Heron's formula states that
A=√[s(s-a)(s-b)(s-c)] where s=(a+b+c)/2.
On the other hand, we can calculate cos C in terms of a, b, c using the cosine rule, calculate sin C from cos C using the Pythagorean identity cos²C+sin²C=1 and the fact that sin C≥0, and finally calculate the area using the formula A=½ab sin C.
Our method is to convert the trignometric formula into a formula just involving the sides, using the cosine rule and the Pythagorean identity, and show we end up with Heron's formula. The argument applies whether we are given the three sides a, b, c or two sides a, b and angle C.
By the cosine rule we have
c²=a²+b²-2ab cos C
2ab cos C=a²+b²-c²
We now square both sides so as to be able to use the Pythagorean identity in the form cos²C=1-sin²C:
(2ab cos C)²=(a²+b²-c²)²
4a²b²cos²C=(a²+b²-c²)²
By the Pythagorean identity
4a²b²cos²C=4a²b²(1-sin²C)=4a²b²-4a²b²sin²C
So
4a²b²sin²C=4a²b²-4a²b²cos²C
Applying our equation above for 4a²b²cos²C we get
4a²b²sin²C=4a²b²-(a²+b²-c²)²
=(2ab)²-(a²+b²-c²)²
=[2ab+(a²+b²-c²)][2ab-(a²+b²-c²)]
=[a²+2ab+b²-c²][2ab-a²-b²+c²]
=[(a²+2ab+b²)-c²][c²-(a²-2ab+b²)]
=[(a+b)²-c²][c²-(a-b)²]
=(a+b+c)(a+b-c)[c+(a-b)][c-(a-b)]
=(a+b+c)(a+b-c)(c+a-b)(c+b-a)
=(a+b+c)(a+b+c-2c)(c+a+b-2b)(c+b+a-2a)
=2s(2s-2c)(2s-2b)(2s-2a)
4a²b²sin²C=16s(s-c)(s-b)(s-a)
Dividing both sides by 16:
¼a²b²sin²C=s(s-a)(s-b)(s-c)
(½ab sin C)²=s(s-a)(s-b)(s-c)
Taking the non-negative square root of each side (noting sin C≥0)
½ab sin C=√[s(s-a)(s-b)(s-c)]
which shows the equivalence of the two formulae for the area of a triangle.
BTW, you can get a nice identity from this - a way to calculate sin A directly from the three side lengths:
As ½bc sin A=√[s(s-a)(s-b)(s-c)], we get sin A=2√[s(s-a)(s-b)(s-c)]/bc
For example, for an equilateral triangle of side 2, we get s=3, s-a=s-b=s-c=1, so
sin A=2√[3(s-a)(s-b)(s-c)]/bc
=2√[3×1×1×1]/2×2
=√3/2, as expected.
@@MichaelRothwell1 ah, this solved the problem that was stuck in my mind for a long time. Thank you for the long and detailed response!
Love your videos man :)
Lmao at 2:59, the Sequeeze theorem.
6:36
Sir, can you make a video of the solutions to the integration bee contest questions?
A calc classic 🔥
Thank you blackpenredpen this was very instructive and enjoyable.
Our whole class was definitely taught the derivative of sin before we were taught the proof behind it, anyway I found it’s better to just analyze it like how the lim x->0 of sin (x)/x=1 then you know the derivative at 0 is 1 and you can find the max at pi/2 so you know the derivative there is 0, from there it’s pretty easy for anyone to see the derivative is cos
Now i really know it already
YOU ARE THE BEST, THANK YOU
Hey Blackpenredpen, i think a good video idea would be you doing any calculus based college entrance exam
Is it strictly necessary to define f(0)=0, or can we just omit it by declaring f(x) is continuous? Clearly the squeeze theorem (sequeeze?) tells us the limit of f(x) as x->0 is 0, so defining x(0)=0 can just be filled in by continuous extension.
Where can I get that derivative poster(canvas- 3:45) from?👀
Checked your site. Are they out of stock?
The limit when x->0 of xsin(1/x)=[sin(1/x)]/(1/x) =sint/t when t->○○, because 1/x=t. We know that |sint|=
Use sampling theorm with Dirac delta
Im the i know it already student and i knew it already
Is there a non-piecewise function that can be used to demonstrate this?
Since f is sandwiched between x² and -x², both of which have zero value and derivative at x=0, you can conclude f'(0) = 0 by squeeze theorem (applied to the definition of the derivative on all three functions).
P.S. Ok that is almost identical to the video, just without cancelling x in the denominator.
Could this be a way to conclude that f'(0)=0, and if so, does this method have a name? kind of a squeeze theorem but for derivatives, maybe?
f(x) is bounded on top by g(x)=x^2 and on the bottom by h(x)=-x^2
Given that both g'(x) and h'(x) are continuous and they are equal at x=0, then that must be the derivative of f(x) at that point.
1) ask those students: why did you come here if you already know it?
2) call those students parrots that spit textbook back at the examiner but have no clue where and how they are going to use it.
Either method works :)
1) "I have to be here"
2) there are two variations: "this is untrue, I understand perfectly well where to use it" or "so what?"
Also insulting your students isn't exactly going to motivate them
What's a real life example of where this would be used? Should start with that.
could you explain why in second Order diff equations with complex roots there Is no i in the solution?Is It included in the constant C?shouldnt It be only real?
You know your Teacher/Prof. is the real deal by how he/she switches 2 markers thats being held on a single hand.
I love teaching this one to students.
Sir it should be 1 right..?? Lt x->0 sin (1/x)/1/x =1 😢😢
Very nice example, great👌
I thought the guys who know the thing already slept through the class
Instead of using the "sequezze" theorem, can we bring the x down down?
6:30 The three sides could have been anything. Does this mean anything is zero if we multiply it with X and want to find the limit at 0?
Only if it satisfies the previous step. And it is exactly what the squeeze theorem says.
In this example looking at the squeeze theorem around 0,
Starting with some interval I, where 0 ∈ I, if you happen to know that
a
I need help with Thomae's function on the interval (0,1). Where is the function continuous? Where does the limit exist? Where is it differentiable?
I’m curious, why is the squeeze theorem necessary to prove that 2xsin(1/x) = 0 for x = 0? Isn’t it obvious that no matter what x is sin is bounded between -1 and 1 and 0 multiplied by anything between -1 and 1 is zero?
I wondered the same thing but I guess math just likes to be rigorous which is fair enough
Yes, the squeeze theorem is just a formal way to show it, but your intuition is right and many mathematicians would also just say it equals 0 and not bat an eye.
But you don't need squeeze theorem:
Theorem: If lim x->0 f(x)=0 and g(x) is bounded, then lim x->0 f(x)g(x)=0.
Proof: Since g is bounded, let M be such that |g| < M. Let e>0. Since lim x->0 f(x)=0, we can find D such that when |x|
“something between -1 and +1” is of course not known but is a limited number so when is multiplied by 0 is not undefined, it MUST BE 0 … the squeeze theorem could not give a different value 🤷🏻♂️
6:56 me when I forget to turn in my completed assignment
Awesome Teaching :)
lim xsin(1/x) = lim sin(1/x)/(1/x)
let u=1/x
lim when u approaches infinity of
sin(u)/u = 0
I was wondering why the limit doesn't give the right answer... until I remembered that f' is simply not continuous
funny name, in France we call this theorem (sequeeze) the constable theorem (théorème des gendarmes)
I have 2 questions. Doesn't this assume that f(x) = x^2 * sin(1/x) as x approaches 0, an assumption with no reasoning? Also, is the first method of checking the derivative from the left and right flawed, or does is this method inconclusive if it says the limit DNE?
The general theorem says:
if you have a continuous function in an interval, such that the derivative always exists expect at most in a point x0. IF the limits of the derivative from left and right both exists and are equal, then the derivative exists also I n x0 and coincides with those limits.
Here there is the big assumption that the function is continoisv not only in x0, but in an interval surrounding it, and that the only non-differentiable point is at most only x0
Essantially it applies to continous piece-wise functions
guys i would like to say that the function is the thumbnail is oscillating therefore discontinuous and in-differentiable. so f’(0) = undefined. LESSSSS GO SON
My Calculus I professor called it the "pinching theorem." He was from India and his accent was difficult for me to understand at times. I thought for a while he was saying "peachy theorem."
Why isnt differentiating at x=0 valid? Surely that’s valid?