I love your calm, methodical, and understandable approach to teaching math skills! Thank you for the time and effort you put into these videos. these are great resources for students.
My dear friend, after 100 years, people will still be watching your videos to learn calculus. You are so involved in your delivery, sharing and excellent in your concepts.
I’m always pleasantly impressed by the love you have to teach mathematical skills, by your large scale knowledge field, by your calm and your pedagogy. By the way, I’ve always studied mathematics in French but I must recognise that I understand every subtility of your langage because your prononciation is agreeable and clear. Thanks so much for the effort you do to make mathematics so abordable and so easy.
If this is a problem to be solved in an exam We will need you to print me at each step. Great solution with logical explanation. Love to learn more with you❤
For anyone who knows the subject, this is a one-liner: log(x^3+1/x^3)=log x^3(1+x^{-6})=3log x+1/x^6+o(x^{-12}). After dividing by x, this tends to 0. By continuity argument, the answer is 1.
I was thinking of a more simplistic approach. As x goes to infinity the term 1/x^3 goes to zero and then the limit could be reduced to lnx^3/x, which translates to 3*lnx/x. As the lnx functions grows slower than any polynomial function the limit goes to zero and therefore the overall limit is 1. Is this way of thinking applicable?
Instead of using L’H rule, we could factor out x^3 inside (x^3+1 /x^3 and then use logarithm and the limite of lnx/x which is 0 when x approaches to the Infinity . So we get the same result.
Hi there, it has a much simpler solution, since as X approaches towards infinity the term 1/x^3, becomes 0, therefore it'll be simplified to find the lim of [(X^3)]^(1/x) when X approaches to infinity. That is equal to lim x^(3/x) , and in turn it is equal the lim of x^0 as X approaches the infinity, therefore it's equal to "1".
6:36 Instead of multiplying by x^4 it is simpler to immediately divide limit by x^2. Then lim x->inf (3-3/x^6)/(x+1/x^5)=3/inf=0. By the way, it avoids the confusion of the 0-0 sign. 4.26 The left limit is unnecessary. lim x->inf ln(y)= ln(y).😎
loved your approach man!btw at 6:25 when you applied L'hopital's rule and realized you needed to use some algebraic manipulation to evaluate the limit, if you kept applying L'Hopital 's rule until it was possible to evaluate the limit would you get the right answer?I mean, it's not that practical, but I'm just curious
You mate are the reason I started to like calculus, and quite frankly you’re the reason I believe I can make it and become an engineer (although a manager one cause I still like money more than math 😂)
Here 1 / x^3 tends to 0. so, result = [ x^(1/x) ]^3 = 1^3 = 1. (1 + zero) ^ infinity = e^constant but infinity ^ zero = 1 in the normal case of x and 1/x being infinity and zero. In this problem, powers of 1/x don't matter since they don't play any material role and all powers of x^(1/x) are powers of 1 = 1. So power of 3 also does not play any role. *General Rule* (X^n + X^-m)^(c/X) will tend to 1 for all positive integers m, n, c with X tending to infinity.
PROFESOR, UN GUSTO SALUDARLE. EN EL MINUTO 6.34 DERIVA EL LOGARITMO, PERO LA X DEL DENOMINDOR NO SE DERIVA ?. POR FAVOR SU COMENTARIO. AGRADECIDO POR LOS VIDEOS. SALUDOS DESDE PERU.
just a question: we can reorder it in a root format, where we get the y=inf'th root of (x^3+1/x^3). since y is not exponential, and (x^3+1/x^3)>=1 when x->inf, i can easily see, that the inf'th root of y is 1 when x->inf. or am i missing something?
The Ln function tends, if we can say, so "weakly" to the infinity, compared to X and actually the x is faster than ln |p(n,x )| for any polynomial of any fixed degree n. And for sufficiently large X , the inequality X> ln | p(n,X)| holds.
A simpler solution is to say as x-> infinite => y = (x^3) ^(1/x) = x^(3/x) from inspection as x-> infinite y=1, if you want to use L"H you can say y=x^(3/x) so ln(y)= (3/x) * ln (x) now do L"H and so you get (1/x) /1 or 1/x as x-> infinite ln(y)=0 => y=1
Quite often mathematics gets to a solution by applying a convenient (not necessarily true) assumption. Here it seems to be an assumption that (x) = (x+3) when x becomes infinite. If you imagine that any weight that we cannot express, measure or weigh is the same or equal to infinite, we make a mistake. A heavier rock is always heavier than a lighter one, no matter how small the difference in weight is.
indeed a nice solution. however wouldn't the same logic be applied to what you said about the limit to the exponential? i rewrote the inside as e^(1/x)(ln(x^3+1/x^3)) and then put the limit into the exponent because exponential of base e if limit exists, exists for all exponents. then evalueate limit. i got zero too via the same method as you. then just replace that with e^0 and then of course you get 1.
I knew that it would be 1 bc x³+1÷x^3 when it's infinity is really just x^3 and bc 1÷x would outgtow x^3 really fast so it's really just x^1÷x and that is 1(when x is aproching infinity)
Can you not solve this limit without use L’Hospital? It is strange because I thought that limit was developed before limit so how Euler solve determine his number??
I don't know but as soon as I realized that the expression is raised to 1/x, I thought in a very very large number raised to that power and to me the answer was 1 similarly as x tends to infinity the limit should be 1.
It's not like it only works for 0/0 or inf/inf but these are the only two situations for which the differentiation would be necessary. If either the denominator or the numerator was neither zero or inifinity then you would have already got the result you were looking for, so you differentiate the expressions on both sides till you get a result as such. E.g if the denominator was zero and the nominator wasn't then clearly the limit would be infinity so there would be no need to further differentiate the equation. You wouldn't need to differentiate an equatoin whose target limit is 2/0 coz you would know the limit would be ∞. L'H rule works for all cases except the target result only practically means something to you when the target value isn't 0/0 or inf/inf.
@@junchen9954 uhmm.. the derivation of lhopital comes by assuming a 0/0 form (or inf/inf). So I don't think that it will apply to other cases. (The proof is in 3b1b calculus playlist if you are interested)
i just learned limits 2 days ago so someone please tell me if i did something horrendously wrong 💀💀💀💀: let y = x^3 + 1/x^3 so lim x -> ∞ (y)^1/x so x for infinity is a number that keeps getting bigger and bigger and (y)^0= 1 so the answer is just 1?
The y approaches infinity which leaves us with an indeterminate form of limit, infinity^0. But honestly tho, you did pretty good for the first 2 day of limits
0^0 is not one always. If you look at the function x^0 it seems 0^0 will be one but if you look at the function 0^x it seems to the value will be 0@@helm36
I would say that it can be easier ... ( X^3 + 1/X^3 ) ^(1/x) the second term in the sum goes to zero ... so left ( x ^ 3 ) ^( 1/x ) ... x ^ ( 3/x ) and then ln en de l hopital is easier ....
Occorre specificare che la variabile reale x ,tende a + infinito ; la funzione a valori reali di cui si vuole calcolare il limite , non è definita per tutti i valori di x negativi.
why do we not use just a calculator or spreadsheet , just to get a feeling ..., that is the way I explain it to my children ...because de l hopital is tricky because of the conditions of it ...
This is an example of a incredibly complicated description of a trivial result. The "+1/x^3" could not possibly make a difference since this quantity goes to zero. The problem reduces to proving (x^3)^(1/x) goes to zero as x goes to infinity. Taking logs we get (3/x)log(x). Everyone knows that Log(x)/x goes to zero as x goes to infinity. So this could be proved in about 30 seconds.
There is no need for all of this, generally anything, to power zero equal to one ,so from the beginning, we have infinity to power zero, which is equal to one
Isn’t it obvious from the beginning, that the limit is 1? If you look at the initial expression, it is clear that it is like “something” to the 0 degree. So, what ever is “inside” the brackets, whole expression is in zero degree. Anything in 0 degree equals 1. 🤔
from quick calculation by head, I also found limit = 1 ..in less than 5 seconds x^3 + 1/x^3 -> x^3 when x -> infinite (X^3)^(1/x) = x^(3/x) 3/x -> 0 and exponential part prevails upon exponented part so -> x^0 = 1 I agree, it's dirty. .. a good formal proof will always remain better.
Ok let me teach you shortcut to find limit to solve easily any question For infinity put x=999999 digits as much you want For 0 put 0.00001 more digits more accurate answer same for other numbers Now if we see this question (9999999³+(1/999999³))^(1/999999) So it becomes (9999999999......)⁰ As we know a⁰=1 Thus answer is 1 This trick almost work for all questions and if it's MCQ question it's fastest way to find your answer
@@265userwell infinity also no so anything power to 0 is 1 But in case of 0⁰ there are different answer Like anything power to 0 is 1 0 to power anything is 0 so it's undetermined
I feel like I'm watching Bobb Rosss but it's the math edition.
Hahaha Or Mr. Rogers does math.
"Over here we have some happy little limits."
Outstanding leszon.
I love your calm, methodical, and understandable approach to teaching math skills! Thank you for the time and effort you put into these videos. these are great resources for students.
I agree :)
It's enjoyable to watch somebody take so much pleasure in doing and teaching math. It makes it so much more interesting!
My dear friend, after 100 years, people will still be watching your videos to learn calculus. You are so involved in your delivery, sharing and excellent in your concepts.
I’m always pleasantly impressed by the love you have to teach mathematical skills, by your large scale knowledge field, by your calm and your pedagogy. By the way, I’ve always studied mathematics in French but I must recognise that I understand every subtility of your langage because your prononciation is agreeable and clear. Thanks so much for the effort you do to make mathematics so abordable and so easy.
Wow, thank you!
sir you explain it very simply, thank you
I found your video randomly and found it that your teaching way is unique and started to watch regularly ❤
Nice handwriting, clear explanation and a smooth pace. Nicely done!
Just found your channel, been loving the videos
Thank you Sir for these videos!
Gosh, this video is so refreshing. I love it.
We will never stop learning
You never stop teaching ❤
ur shirt is definitly amazing i like it and your way of explaining
Thank you so much 😀
The world’s most lovable math teacher! The anti-Nash.
If this is a problem to be solved in an exam
We will need you to print me at each step.
Great solution with logical explanation.
Love to learn more with you❤
For anyone who knows the subject, this is a one-liner: log(x^3+1/x^3)=log x^3(1+x^{-6})=3log x+1/x^6+o(x^{-12}). After dividing by x, this tends to 0. By continuity argument, the answer is 1.
Hello from Brazil, like ur channel very much I’m civil Engineering student 🤝
I was thinking of a more simplistic approach. As x goes to infinity the term 1/x^3 goes to zero and then the limit could be reduced to lnx^3/x, which translates to 3*lnx/x. As the lnx functions grows slower than any polynomial function the limit goes to zero and therefore the overall limit is 1. Is this way of thinking applicable?
You have made my work easier sir 😁😉
Instead of using L’H rule, we could factor out x^3 inside (x^3+1 /x^3 and then use logarithm and the limite of lnx/x which is 0 when x approaches to the Infinity . So we get the same result.
What kind of teacher are you? 😮 Your teachings are always amazing ❤🎉 Keep it up.❤
Hi there, it has a much simpler solution, since as X approaches towards infinity the term 1/x^3, becomes 0, therefore it'll be simplified to find the lim of [(X^3)]^(1/x) when X approaches to infinity. That is equal to lim x^(3/x) , and in turn it is equal the lim of x^0 as X approaches the infinity, therefore it's equal to "1".
No! For istance lim(x→∞)((1+1/x)^x =e not equal to lim(x→∞)(1)^x=1
@@salvatorecosta875The indeterminate form is 1^∞, but not ∞^0
unfortunately that is an indeterminate form can't have ∞^0
This is so well explained! Thank you so much 😇
Can't believe that I solved that completely on my own and did everything right!
WELL DONE PRIME NEWTON BRAVO !
loved your infinity t shirt
6:36 Instead of multiplying by x^4 it is simpler to immediately divide limit by x^2. Then lim x->inf (3-3/x^6)/(x+1/x^5)=3/inf=0. By the way, it avoids the confusion of the 0-0 sign.
4.26 The left limit is unnecessary. lim x->inf ln(y)= ln(y).😎
loved your approach man!btw at 6:25 when you applied L'hopital's rule and realized you needed to use some algebraic manipulation to evaluate the limit, if you kept applying L'Hopital 's rule until it was possible to evaluate the limit would you get the right answer?I mean, it's not that practical, but I'm just curious
Clear and excellent job !
Is academic crush a thing? Cuz i think i have one of this guy
Your are the best sir
Nice, like the way you do the maths, and your enthusiasm ❤🎉😊
the shirt is FYE!!! 🔥🔥
Koszulka jest świetna. Z chęcią zakupię.
Love your work!
You mate are the reason I started to like calculus, and quite frankly you’re the reason I believe I can make it and become an engineer (although a manager one cause I still like money more than math 😂)
Here 1 / x^3 tends to 0. so, result = [ x^(1/x) ]^3 = 1^3 = 1.
(1 + zero) ^ infinity = e^constant but infinity ^ zero = 1 in the normal case of x and 1/x being infinity and zero.
In this problem, powers of 1/x don't matter since they don't play any material role and all powers of x^(1/x) are powers of 1 = 1. So power of 3 also does not play any role.
*General Rule* (X^n + X^-m)^(c/X) will tend to 1 for all positive integers m, n, c with X tending to infinity.
Great explanation ❤ , please make videos on convergence and divergence of infinite series
PROFESOR, UN GUSTO SALUDARLE. EN EL MINUTO 6.34 DERIVA EL LOGARITMO, PERO LA X DEL DENOMINDOR NO SE DERIVA ?. POR FAVOR SU COMENTARIO. AGRADECIDO POR LOS VIDEOS. SALUDOS DESDE PERU.
Never stop learning to infinity!!!
explained very well. Thank you.
Very good example great👍
Where can I get this T-shirt?
Working on it
Everyting raised to the power of 0 IS one 🎉🎉
just a question:
we can reorder it in a root format, where we get the y=inf'th root of (x^3+1/x^3). since y is not exponential, and (x^3+1/x^3)>=1 when x->inf, i can easily see, that the inf'th root of y is 1 when x->inf.
or am i missing something?
Very elegant
The Ln function tends, if we can say, so "weakly" to the infinity, compared to X and actually the x is faster than ln |p(n,x )| for any polynomial of any fixed degree n. And for sufficiently large X , the inequality X> ln | p(n,X)| holds.
You are doing well !!
11:50 almost like this limit ultimately has won
This is brilliant
I liked your shirt.
Very good. Thanks 🙏
You may write e^y at ðe beginning so you don't forget. Just in case
A simpler solution is to say as x-> infinite => y = (x^3) ^(1/x) = x^(3/x) from inspection as x-> infinite y=1, if you want to use L"H you can say y=x^(3/x) so ln(y)= (3/x) * ln (x) now do L"H and so you get (1/x) /1 or 1/x as x-> infinite ln(y)=0 => y=1
Quite often mathematics gets to a solution by applying a convenient (not necessarily true) assumption. Here it seems to be an assumption that (x) = (x+3) when x becomes infinite. If you imagine that any weight that we cannot express, measure or weigh is the same or equal to infinite, we make a mistake. A heavier rock is always heavier than a lighter one, no matter how small the difference in weight is.
indeed a nice solution. however wouldn't the same logic be applied to what you said about the limit to the exponential? i rewrote the inside as e^(1/x)(ln(x^3+1/x^3)) and then put the limit into the exponent because exponential of base e if limit exists, exists for all exponents. then evalueate limit. i got zero too via the same method as you. then just replace that with e^0 and then of course you get 1.
Nice one. can you take the limit as x approaches 0? The result of that is more fun and unexpected
Sir, I'm in love with you!
Me too 😆
Hey can you have a look at this
lim(x->♾️)[4^n+5^n]^1/n
?
I like this
@@PrimeNewtons then how about a video on it
I knew that it would be 1 bc x³+1÷x^3 when it's infinity is really just x^3 and bc 1÷x would outgtow x^3 really fast so it's really just x^1÷x and that is 1(when x is aproching infinity)
Actually perfect.
Can you not solve this limit without use L’Hospital? It is strange because I thought that limit was developed before limit so how Euler solve determine his number??
I don't know but as soon as I realized that the expression is raised to 1/x, I thought in a very very large number raised to that power and to me the answer was 1 similarly as x tends to infinity the limit should be 1.
Wow it's so good
I transformed it into and exponential and then I neglected the 1 before the x^6 and I got the same result. Is it right ?
I am solve limit any questions by Infinite Countability 😊. Some times I am wrong but most of the time my right.
How did you use Lhopital? It only works when you have a 0/0 indeterminate form right?
But we will still get 1 if we use the property of logs + it's taylor expansion to solve it
I used to think so too, but 0/0 is also inf/inf. So it works for both. Never Stop Learning!
@@PrimeNewtons cool! And thanks
It's not like it only works for 0/0 or inf/inf but these are the only two situations for which the differentiation would be necessary. If either the denominator or the numerator was neither zero or inifinity then you would have already got the result you were looking for, so you differentiate the expressions on both sides till you get a result as such. E.g if the denominator was zero and the nominator wasn't then clearly the limit would be infinity so there would be no need to further differentiate the equation. You wouldn't need to differentiate an equatoin whose target limit is 2/0 coz you would know the limit would be ∞. L'H rule works for all cases except the target result only practically means something to you when the target value isn't 0/0 or inf/inf.
@@junchen9954 uhmm.. the derivation of lhopital comes by assuming a 0/0 form (or inf/inf). So I don't think that it will apply to other cases. (The proof is in 3b1b calculus playlist if you are interested)
Don't we immediately see that a parenthesis to the zeroth power will be equal to one?
🙏 Thank you Sir
but sir where is the natural log taken at the function
All I can say is: Wow.
Substitute a large valve for x ,Example x=1000
You will get the answer 1
If I draw a graph of that function I see the value of 2 as the limit goes to infinity. Where is the mistake?
I was wrong - must go farther than some hundred to see the Limit of 1...
Simply substitute x as infinity.knowing that infinity to the power of 3 . It equals( 2)^1/1.
This is 2.
i just learned limits 2 days ago so someone please tell me if i did something horrendously wrong 💀💀💀💀:
let y = x^3 + 1/x^3
so lim x -> ∞
(y)^1/x
so x for infinity is a number that keeps getting bigger and bigger
and (y)^0= 1
so the answer is just 1?
The y approaches infinity which leaves us with an indeterminate form of limit, infinity^0. But honestly tho, you did pretty good for the first 2 day of limits
Is it wrong to just raise infinity to zero and say its 1? Since any numbers raised to zero is 1
Nah, infinity is not a number. Also the case of anything to the zero being 1 isn't always true as 0^0 is a counterexample
What?0^0 is 1, it's not a counterexample
0^0 is not one always. If you look at the function x^0 it seems 0^0 will be one but if you look at the function 0^x it seems to the value will be 0@@helm36
@@Kraken-lm1cx function x^x as x approaches 0 equals 1. For example you take 10e-10 to the power of 10e-10 it will be pretty close to 1
@@Kraken-lm1cx yeah 0^x as x approaches 0 is not 1, but 0^0 being 1 is basically a math rule due to the reason I mentioned above
But u can tell that limit approaches 1 from the exponent. Because the exponent is 0.
I would say that it can be easier ... ( X^3 + 1/X^3 ) ^(1/x) the second term in the sum goes to zero ... so left ( x ^ 3 ) ^( 1/x ) ... x ^ ( 3/x ) and then ln en de l hopital is easier ....
Occorre specificare che la variabile reale x ,tende a + infinito ; la funzione a valori reali di cui si vuole calcolare il limite , non è definita per tutti i valori di x negativi.
Anynumber to the power of 0 is 1?
Using ln(ab) and L.H. , gives
lnY=0, and so Y=1 as x→∞
why do we not use just a calculator or spreadsheet , just to get a feeling ..., that is the way I explain it to my children ...because de l hopital is tricky because of the conditions of it ...
This is an example of a incredibly complicated description of a trivial result. The "+1/x^3" could not possibly make a difference since this quantity goes to zero. The problem reduces to proving (x^3)^(1/x) goes to zero as x goes to infinity. Taking logs we get (3/x)log(x). Everyone knows that Log(x)/x goes to zero as x goes to infinity. So this could be proved in about 30 seconds.
There is no need for all of this, generally anything, to power zero equal to one ,so from the beginning, we have infinity to power zero, which is equal to one
Are you sure?
@@PrimeNewtons why not ?
I love this
Beautifully done
It is much easier to do without L'Hopital's rule. Through the sum of cubes
The result I found in 6 sec.
Nice
infinity to the power of zero is 1
Isn’t it obvious from the beginning, that the limit is 1? If you look at the initial expression, it is clear that it is like “something” to the 0 degree. So, what ever is “inside” the brackets, whole expression is in zero degree. Anything in 0 degree equals 1. 🤔
Try graphing it desmos and you will see that the limit is 0 and not 1.
from quick calculation by head, I also found limit = 1 ..in less than 5 seconds
x^3 + 1/x^3 -> x^3 when x -> infinite
(X^3)^(1/x) = x^(3/x)
3/x -> 0 and exponential part prevails upon exponented part so -> x^0 = 1
I agree, it's dirty.
.. a good formal proof will always remain better.
Please am confused 😢
Already 2025 and still struggling
Let x = 99999999
And you will get
1.000000553
Approx. = 1
Ok let me teach you shortcut to find limit to solve easily any question
For infinity put x=999999 digits as much you want
For 0 put 0.00001 more digits more accurate answer same for other numbers
Now if we see this question
(9999999³+(1/999999³))^(1/999999)
So it becomes (9999999999......)⁰
As we know a⁰=1
Thus answer is 1
This trick almost work for all questions and if it's MCQ question it's fastest way to find your answer
I'll make sure to use this in exams
of course this is great for mcqs but when it comes to a 6 marker written method then you will unfortunately have to learn how limits actually work 😂
He said infinity raised to zero it's undefined... Can someone clarify this am lost
@@265userwell infinity also no so anything power to 0 is 1
But in case of 0⁰ there are different answer
Like anything power to 0 is 1
0 to power anything is 0
so it's undetermined
@@SpicyGregPowersyaah but you can use it to cross check your answer
But in MCQ it is very helpful
easy
just get rid of 3/x^4 and 1/x^3 beacause x goes to infinity then theses fractions is 0 then you can simplify by 3x^2/x^3=3/inf=0
The dislike button is not enough
let k= 1/x so the L= lim k--> 0 (k³+1/k³)^k = 1.