Curb your L'hopital's Rule

Рет қаралды 12,669

Күн бұрын

Пікірлер: 69

@michellaboureur7651 11 ай бұрын

I wish I’d had such a maths teacher, you’re so considerate and benevolent. Having pupils feel loved and cared for is the first essential element in pedagogy. Not for the sake of kindness alone but because of what that means about teacher’s ability to understand the needs of pupils. The other element of course is competence in the subject matter. You are endowed with both.

@ciprianteasca7823 11 ай бұрын

All your comments raised to the power of...infinity!

@dan-florinchereches4892 4 ай бұрын

I am wondering why he did the change of variable. My first thought was to factor out forcefully 9^x out of denominator and nominator and have (8/9)^x and (7/9)^x go to 0 as X grows large. Then I noticed the limit is towards -inf so just factor out 7^x and have to quantities >1 go to zero as exponent grows smaller. I like his voice and would like to understand his method of teaching because as I am now if I wanted to be a teacher I guess I would epically fail to make students understand my love of maths...

@Zerotoinfinityroad 11 ай бұрын

One of the bestest teachers I've Ever seen😇

@Moj94 11 ай бұрын

I can confirm that I was throwing that at every limit I could find. :)) My L'Hopital brain would rather skip the question some years ago.

@DatBoi_TheGudBIAS 11 ай бұрын

as a person who learned to do limits like these by head quite fast with the "the greatest matters more" rule, i immediatly saw -1 as the answer

@didar8809 11 ай бұрын

Just the best teacher

@BartBuzz 11 ай бұрын

Excellent! Sometimes we forget about the basics!

@gastonsolaril.237 11 ай бұрын

Amazing video, as always!! Just a last-minute idea: I believe that another interesting (though messy) way to solve these, is through Taylor series... in theory, "a^x = e^(x ln a)". Perhaps, in the end, you have 3 power series above and below, and you could join them. All of them have the same max-degree term approaching infinity at the same pace, so you may end up with a typical infinity/infinity case which when approached symbolically, may end up with the same right result!

@jensberling2341 11 ай бұрын

Thank you. How I love your presentation

@hqs9585 11 ай бұрын

4:46. Did you make a mistake with the signs in the denominator?

@KundrahOmusamaHerbert 10 ай бұрын

What a such interesting content this is!

@emmanuelbossfx 7 ай бұрын

I saw that mistake From the beginning but thank God 🙏 you discovered it

@punditgi 11 ай бұрын

Prime Newtons does it all! 🎉😊

@vishalmishra3046 3 ай бұрын

7 < 8 < 9 so their negative powers are opposite, 1/7 > 1/8 > 1/9. So, turn this into a positive infinity limit problem by dividing by the largest term 7^x or (1/7)^(-x) same thing. Numerator tends to 7^x (nearly zero) and denominator tends to -7^x (nearly negative zero) but they are both equal in magnitude and opposite in sign. So *result = -1*

@ChadTanker 11 ай бұрын

But it's symmetrical... so you could just go ahead and rewrite the fraction so the top and bottom lines up. lim x-> -inf. ( (9^x - 8^x + 7^x) / (9^x + 8^x - 7^x) ) And then you cancel like terms by simply dividing leaving: lim x->inf. ( 1 - 1 - 1) which is just 1 - 2 = -1 way easier and quicker without much thaught.

@chaosredefined3834 9 ай бұрын

You can't cancel terms like that. Suppose we have (a - b + c)/(a + b - c), by what you just did, we get 1 - 1 - 1 = -1. But if I put in a = 2, b = 6, c = 10, we get 6/-4 = -1.5, but by your logic, we have -1.

@rashidissa5887 4 күн бұрын

Still in problem with the"limit" issue but hopefully, by following your classes, I will one day catch up

@therichcircle.8819 11 ай бұрын

Best tutor, you are so lovely

@rajesh29rangan 9 ай бұрын

Elegant solution.

@zakariakhalifa9681 11 ай бұрын

Just awesome

@kennethgee2004 11 ай бұрын

You could have done all the simplifying first and then did the change of variable it necessary. I also saw that everything was in terms of a^x, such that taking the ln of the terms to pull out x to the front would have been an easier way to approach this.

@SuperTommox 11 ай бұрын

Gotta give love to the algebra before you give it to calculus!

@Archimedes_Notes 11 ай бұрын

Assume now that we are facing the same original problem but we are taking the limit toward positive infinity; what would be the limit?

@fredfred9847 11 ай бұрын

@Archimedes_Notes 11 ай бұрын

That is what i got Gracias

@אסףרביבו-ר7ת 8 ай бұрын

But you didnt solve for x but for t still you a great teacher and a person i love you man❤

@TheFrewah 7 ай бұрын

He did because the end result doesn’t depend on x

@Theterrorlogs-0 11 ай бұрын

Its funny how you at first did easy questions and now hard ones

@surendrakverma555 11 ай бұрын

Very good. Thanks 🙏

@jumpman8282 11 ай бұрын

Sneaky! I exhausted pretty much every algebraic trick in the book, and even tried L'Hôpital's rule as a second-to-last resort, before realizing that all I had to do was think about dominant terms. Even then, I thought 9^𝑥 was the dominant term, so I divided everything by 9^𝑥. But about halfway through, I realized that since 𝑥 is approaching _negative_ infinity it's actually 7^𝑥 that is the dominant term. And sure enough, dividing everything by 7^𝑥, the problem basically solved itself. Oof.

@sadeqirfan5582 9 ай бұрын

You could just factorise: numerator = - denominator. Cancels out and is -1.

@cribless810 11 ай бұрын

TitIe got me cIicking immediateIy🤣🤣

@DonutOfNinja 11 ай бұрын

You can also use l'hopital 7 times, ie taking the 7th derivative of both sides and getting a limit that can be simplified to 7!/(-7!)

@KRO_VLOGS 11 ай бұрын

Sir can you make a video of differentiating general x^n using first principal

@Orillians 11 ай бұрын

he did!

@KRO_VLOGS 11 ай бұрын

@@Orillians can't find it

@Orillians 11 ай бұрын

wait your riht. Sorry. My mistake.@@KRO_VLOGS

@klementhajrullaj1222 11 ай бұрын

Division up and down with 7^x

@varun3282 11 ай бұрын

I tried expansions it didn't work out

@m.h.6470 11 ай бұрын

Good that you found the +/- error... that would have messed up the result. 😉

@brunoporcu3207 11 ай бұрын

Bravissimo professor!!!!

@JourneyThroughMath 11 ай бұрын

Im proud of myself😊. I saw Lhopitals rule wouldnt work. So i tried the ration function approach. My only mistake was I multiplied by 1/9^x (I didnt transition to t) instead of 1/7^x. But thats an easy mistake to fix

@nharvey64856 11 ай бұрын

Well done

@naorbedinheinrichm.5167 11 ай бұрын

lezgo prime newton

@jamal369 11 ай бұрын

40 sec ago

@luxxulyanite 11 ай бұрын

This video should have been called "Curb your L'Hospital's rule"

@PrimeNewtons 11 ай бұрын

I agree

@luxxulyanite 11 ай бұрын

@@PrimeNewtons Solved by Larry David 😇

@TSR1942 11 ай бұрын

Damn smart guy.

@alejandropulidorodriguez9723 11 ай бұрын

splendid

@SilasKiprotich-d4r 3 ай бұрын

Clear

@luisangel25 11 ай бұрын

"those stop learning, stop living"

@mikefochtman7164 11 ай бұрын

Well.... I was replacing terms with e. Something like e^(xln(7))+e^(xln(8))-e(xln(9)) and getting no where. That wasn't pretty. lol

@comrade_marshal 11 ай бұрын

Me who forgets about L'Hopital everytime, this time: 🦍🦍🦍 Another reason that L'Hopital won't work according to me is that n^x's derivative returns n^x*ln(n) so, the derivative function is technically nearly similar

@jumpman8282 11 ай бұрын

LOL, I tend to forget STEP ONE, which is to apply direct substitution. I've lost count of the times I found myself lost in a jungle of algebra, just to realize that all I needed to do was "plug it in". They say we learn from our mistakes. Well, I guess this is my personal exception to that rule :)