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.
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! 👍
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.
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.
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.
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 😮
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
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 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
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.
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
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
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.
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
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 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.
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
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.
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!
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)
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 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
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.
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...
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?
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
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
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.
4:46 ...but why? You have a value approaching zero multiplied by an indeterminate value between -1 and 1. What's wrong with assuming the result will approach zero as well?
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?
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.
Hey there, in the small chance that you will see my comment, at 3:40, I am use to seeing the definition of a derivative as limiting h to 0 instead of limiting x to a. Is it possible to use that definition of a derivative? I tried it on this function but it doesn't seem to work without redefining some variables. Thank you
UPDATE: never mind lol, I just tried it again and realised I was using the main expression for f(0) instead of 0. Yes it's easy. That answers my question aha
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."
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.
i mean you could always just graph sin(x) and then draw the derivative by hand and i think at this point it should be really clear that it is just cos(x)
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.
“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 🤷🏻♂️
