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’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.
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.
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❤
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?
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".
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.
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
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.
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?
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 😂)
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).😎
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 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 ....
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)
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.
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
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 ...
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
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
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. 🤔
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.
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
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.
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.
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.
