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 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...
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!
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*
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.
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.
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.
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
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.
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
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 :)
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.
All your comments raised to the power of...infinity!
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...
One of the bestest teachers I've Ever seen😇
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.
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
Just the best teacher
Excellent! Sometimes we forget about the basics!
Thank you. How I love your presentation
I saw that mistake From the beginning but thank God 🙏 you discovered it
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!
Elegant solution.
Just awesome
What a such interesting content this is!
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*
Best tutor, you are so lovely
Gotta give love to the algebra before you give it to calculus!
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.
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.
Very good. Thanks 🙏
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❤
He did because the end result doesn’t depend on x
You could just factorise: numerator = - denominator. Cancels out and is -1.
4:46. Did you make a mistake with the signs in the denominator?
Division up and down with 7^x
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!)
Well done
Good that you found the +/- error... that would have messed up the result. 😉
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.
splendid
Damn smart guy.
Bravissimo professor!!!!
Its funny how you at first did easy questions and now hard ones
Clear
Sir can you make a video of differentiating general x^n using first principal
he did!
@@Orillians can't find it
wait your riht. Sorry. My mistake.@@KRO_VLOGS
Assume now that we are facing the same original problem but we are taking the limit toward positive infinity; what would be the limit?
1
That is what i got Gracias
lezgo prime newton
40 sec ago
This video should have been called "Curb your L'Hospital's rule"
I agree
@@PrimeNewtons Solved by Larry David 😇
"those stop learning, stop living"
NICEEEE
I tried expansions it didn't work out
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
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 :)
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
GREAT PROBLEM :) :) :)
Niceeeee
The limit is wrong. Let us observe the denominator -7^(-t). He transformed it to +(1/7^(t)).
You think it's wrong or you know it's wrong?
In the end he corrected it, even if it wasn’t corrected so what? We did learn the thinking process which is the main goal, isn’t it?
Hi again
Man i applied L hospital rule and got stuck 😅
Answer = 1 is it (Spammer)
t = -1 , x = 1
niceeee hahahaa
TitIe got me cIicking immediateIy🤣🤣