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.

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.

One of the bestest teachers I've Ever seen😇

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

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

Just the best teacher

Excellent! Sometimes we forget about the basics!

Thank you. How I love your presentation

Prime Newtons does it all! 🎉😊

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!

What a such interesting content this is!

Very good. Thanks 🙏

Just awesome

Elegant solution.

Best tutor, you are so lovely

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

Well done

splendid

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.

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

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

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!)

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

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.

Damn smart guy.

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

NICEEEE

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

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

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

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

Division up and down with 7^x

lezgo prime newton

I tried expansions it didn't work out

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.

Niceeeee

GREAT PROBLEM :) :) :)

TitIe got me cIicking immediateIy🤣🤣

Bravissimo professor!!!!

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

niceeee hahahaa

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

"those stop learning, stop living"

Hi again

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

Man i applied L hospital rule and got stuck 😅

The limit is wrong. Let us observe the denominator -7^(-t). He transformed it to +(1/7^(t)).

Answer = 1 is it (Spammer)