Creative Solution I should say, In fact the manipulation of -Sin T + Sin T could be used this way makes this video so special. The objective is not to use L'Hôpital and this solution is excellent. The moral lesson for this video is just keep innovating. I have watched a solution to this question without L'Hôpital rule and it is just too complex. Keep up the good work.
I really like when you have a problem and you solve it using different methods. What you learn is that there’s more than one tool in your toolbox. Some tools may be better depending on the problem.
I like the phrase you say at the end of the videos "Those who stop learning stop living", I am a student right now but I will keep learning even after i finish highschool and college. I won't have a career in mathematics probably but the concept is very beautiful, I like what you are doing. Have a good day
I love to use Taylor series rather than Lhopital. It's just feels good to do so for some reason Edit: just completed the video. WOAH THAT WAS EPIC!!!!!!!!!!!!!! This limit was 🔥
I mean, isn't L'H just an application of Taylor series? If we use first-order Taylor's to approximate f(x) ~ f(0) + f'(0)*x and g(x) ~ g(0) + g'(0)*x where f(0) = 0 and g(0) = 0 then f(0) / g(0) = [0 + f'(0)x] / [0 + g'(0) * x] = f'(0) / g'(0).
You just need to use L'Hôpital's rule, which involves successively taking the derivative of the numerator and the denominator (x-sinx)'/x^3' --> (1-cosx)'/(3x²)' --> (sinx)'/(6x)' --> cosx /6 --> 1/6 when x-> 0
That’s true, L’Hopital would work. But it is really interesting to use different methods - I would consider this a more first-principles approach and the algebra is really interesting!
The shortest method to the above solution is by applying L'hopital's rule. Differentiate the numerator and denominator you have 1 - cosx and 3 x square. Substitute 0 for x, undefined. Differentiate the second time again, you have 0 + sin X and 6x. Substitute 0 for x undefined. Differentiate the third time, you have cos X and 6, substitute 0 for x and you have 1/6. Maclaurin series is a long method Excellent teaching from you
I solved in a similar way, maybe with fewer manipulations, with the x=3T substitution, and then use sin3T = 3sinT - 4sin^3 T.Then the limit is 3(T-sinT)/27T^3 + 4/27 (sinT/T)^3 => L = (3L+4)/27 => 24L = 4 => L= 1/6
using L'Hôpital's rule just makes this problem a piece of cake (in this case, limit of double differentiation of the function is the same as limit of the function)
Love your videos. Keep it up, my man! Edit: it would be interesting to see if this could be generalised (with clever variable substitution) so that could get the nth Taylor series term in the limit!
Or you can easily [ in the third line ] keep a "T" for "sinT" in the denominator and a "T" squared for (1-cosT) in the denominator Because: Lim sin(x)/x =1 (x--->0) Lim (1-cos(x))/x^2 =1/2 (x--->0)
@@PrimeNewtonsI don't know whether this is what you asked but the 2nd limit can be proved by the 1st limit and the fact that 1-cosx = 2sin²(x/2) and the 1st limit: (1-cosx)/x² = 2sin²(x/2)/x² = 2x²/4x² = (1/2)
That is neat using equation manipulation points of the adding by net zero to maintain equality, grouping two parts by parentheses and multiplying by net "1" equality. It is so strange the 2T substitution for x nade all that manipulation work out. You found three ways to solve this problem: Taylor Series, L'Hopital's Rule and substitution by 2T. Nested in substitution of x = 2L there were: #1) add net zero sin(T) - sin(T), #2) use the double angle formula, #3) separate a sum of the same denominator, #4) multiply one by net "1" identity, #5) use the initial limit definition of L, and then #6) the limit result of multiplied two limits (to evaluate further with adding, subtracting and multiplying only). So each one was 1/6th of a different concept used to find this limit by algebraic manipulations that ironically were the portion of other manipulation steps used after substitution and simpler addition, subtraction and multiplication general algebra. 😂🤣
this is the type of questions,we jee aspirants practice for boosting our confidence, in my school we have to answer this type of questions by Just seeing this with in 15 seconds.
این استاد ریاضی بسیار عالی و زیبا این مسئله را حل کرد ابتدا بسط sin x را نشان داد و سپس انرا بصورت قابل فهم برای یک دانش اموز دبیرستانی حل کرد من از کشور ایران هستم 🎉 🎉 🎉 🎉
The problem is you just assumed that the limit exist. Then you proceeded to find it using this existence. You have to be careful when "solving" for the limit.
The equation trick is a cute one, I saw it in a Serbian book by Mitrinovic from 1960s. You still need to show that the finite limit exists, ie that the infinity is not a solution.
if you didnt knew that, there is something called l´hopital, if you didnt know that. is basically to derivate numerator and denominator of a fraction when it is 0/0 or infinity/infinity
The Taylor series expansion of sinx about x = 0 is x - x³/3! + x⁵/5! - x⁷/7! + O(x⁹) So, x - sinx = x³/3! - x⁵/5! + x⁷/7! + O(x⁹) So, (x - sinx)x³ = 1/3! - x²/5! + x⁴/7! + O(x⁶) Clearly, 1/3! - x²/5! + x⁴/7! + O(x⁶) --> 1/6, as x --> 0.
Dear teacher , we are using some of your problems for our algebra 2 and trig students , in brief version , without explanation , for practice after each lesson, can we use the same problems or we should change them not to be identical , please advise 🎉
Another, simpler method with the same substitution is at 6:16 you can apply L'hopitals rule straight away to end up with (1 - cos(2T)) / (12T^2) after some manipulations. Then you can apply L'hopitals again to get sin(2T) / 12T. Finally, you can apply L'hopitals rule a third time to reach cos(2T) / 6 -> 1 / 6.
It is very easy if you use the Taylor: sin (x) = x - 1/6 x^ 3 +- You obtain : (x- sin*x)/ x^3 = 1/6 - + term with x^ n in denominator therefore 1/6 is limit
The presenter said and showed this up front in the video. Then asked (paraphrased), "What if the student is in Calc I, where he hasn't come across Taylor series yet?" [In American schools, Taylor series don't get taught til 2nd semester Calculus] Anyway, the development of Taylor series requires the use of derivatives. If you can do that, you might as well just use L'Hopital's Rule and get the answer without bringing in a memorized Taylor series for sin(x). Personally, I like knowing & usiong general techniques (like L'Hop's Rule) that can cover a whole bunch of different situations rather than memorizing a bunch of formulas that that are either handed to you without understanding, required **PERFECT** memorization (off by one coefficient or a sign, and the entire thing's blown), and/or is limited by specific forms that may not fit the problem at hand.
... Chèr ami Newton, Quel bel hommage vous avez rendu à cette limite qui m'est encore précieuse au sense du Calcul(us) merveilleux ! Je devais juste te dire ceci en francais (ma langue maternelle), ci cela ne te dérange pas (lol) ... I gave you some homework friend Newton to try to translate and understand these few words in French; I assure you, only positive ones ... a Newtonian presentation to remember for a long long time my friend, thank you for this great effort (the reason why I love math) ... only the best, Jan-W p.s. If you have some spare time, look at this limit I found in my little archive ... LIMIT(X -> inf. ) [ (X + COS(X) ) / X ] or LIMIT(X -> inf. ) [ (X + SIN(X) / X ] ... and try to explain (why or why not ?) how you would solve this one ... I'm curious, so good luck Newton only when you find the time, okay ? .... thanks again ....
@@samueldeandrade8535 ... Good day Samuel, With these ' indeterminate ' limits I only wanted to show for people who directly jump into " L'Hôpital's Rule ", that this rule fails in this case due to the oscillating SIN/COS function terms (when X goes to infinity) in the numerator, that's all ... your answer is perfectly correct ! ... thank you for your comment ... Jan-W
Taylor expanding is a better way as any high schooler can do it, most people don’t know L’Hôpital, and out of the ones that do know it, I guarantee less than half know the regularity conditions that f and g need to satisfy in the indeterminate limit f/g
Why couldn't we use L'Hopital's rule? [x-sin(x)]/x^3 becomes [1-cos(x)]/3x^2 [1-cos(x)]/3x^2 becomes sin(x)/6x sin(x)/6x becomes cos(x)/6, which approaches 1/6 as x approaches 0.
No, it's not. Because as the instructor says, you are first making an ASSUMPTION that the limit exists, and based on that premises you calculate the limit.. The problem is that with this method is that you can get extraneous solutions for L. But I do agree the algebra used to arrive at the answer is cool nonetheless.
Why not just use the power series for sin x? So subtracting from x, you get x^3/6 - higher order terms which -> 0 as x -> 0, so the limit is immediately seen to be 1/6. This sort of thinking is actually very valuable in the world of engineering.
If I graph that limit function I get a kind of very small hyperbola, where apparently you can see that its limit is at 1/6 (which is the center), what happens there? What happens if instead of dividing by x^3 it is divided by |x|^3? If you graph it and dissect it by lateral limits on each of the values for the absolute value you should notice that they tend to - ∞. How is that possible?
I feel like I understood every step pretty much but I wouldn't have known which overall path to take to find the solution (for the second method especially). How did you 'see' that inserting the "-sin(t)+sin(t)" and factorising would produce a clone of the original limit?
You are correct. This sort of algebra just makes maths look like magic, which turns off countless people from ever learning it. Sure, if you are smart, you might feel "privileged" and "special" because you can understand it and maybe even able to apply this method, but simpler brut force methods are more often than not applied to problems like this like turning sin(x) into a Taylor series expansion about x = 0 and the answer drops out after just a few lines, or immediately in your head if you are used to dealing with power series.
To add on to David's comment, a lot of the problems with unusual solutions like this were found by figuring out a cool trick first and working backwards to make up a problem where it can be applied. No one is expected to notice things like this with consistency.
The way to see it is to want to find another L in your formula. You have a T / T^3 and you want (T - sin T) / T^3 so you insert a - sin T. Then you have to add a + sin T to make it right again. So far you're just exploring options, but then you see that the + sin T you have to add combines nicely with the rest of the formula because you can factor it out. So that's a promising avenue and you go that way. You won't have certainty that this trick will work. Solving a problem often means trying lots of approaches and following the ones that work.
I noticed it too, it’s because it’s not usually proved unless u take analysis, and even then that is under the assumption of many regularity conditions which need not hold, the general result isn’t really proven to students and so if u want full satisfaction, try evaluate limits without using it. Obviously in these examples, the functions involved are everywhere analytic and the regularity assumptions hold, so it’s not so much a problem here
The easiest way is to use the Taylor expansion x-sinx = x- (x-x³/3! + R(x⁵) = x³/6 + R( x⁵). So after deciding by x⁶ you get f(X) = 1/6 + R(x²) which tends to 1/6.
Exactly! This is the right method, as it is done from first principles. People using L’Hôpital are just applying a formula without understanding why it works, when using Taylor expansions are the only reason L’Hôpital works in the first place
@@adw1z Yeah, and actually I use it in my brin to find the solution in a intuitive way, but the problem is that L’Hôpital rule can be used as many times as you can can differentiate the functions, while the Taylor expansion method requires that the function be of class C∞.
Good day eddhairrykrokmou, ... LIM(X -> 0) [ SIN(X) / X ] = 1 (SQUEEZE THEOREM ... remember? ) ... now , LIM(X -> 0) [ ( SIN^3(X) / X^3 ] = LIM(X -> 0) [ (SIN(X) / X)^3 ] = ( LIM(X -> 0) [ SIN(X) / X ] )^3 = 1^3 = 1 ... I hope this helps you out ... just apply all the usual limit laws you were taught! p.s. By the way, lim sin(T^3)/T^3 is NOT the right notation in your question, but lim sin^3(T)/T^3 is !
