@@dutchie265 não é o dobro. quando se iguala as bases fica 3(1+3t)+3= 3+9t+3=6+9t. ele se confundiu porque por coincidência 2*2= 2². mas a generalização de d/dt [ ln(1+at) + at/(1+at)] = (2a+a²t)/(1+at)². ele pensou em (2a+2at)/(1+at)² erroneamente.
This is one of those "oh yeah!" Problems. I was never assigned anything like this in my calculus days. Nor have I assigned anything like this. So I looked at this problem and didnt even know where to start. But every step of the way, my response was "oh yeah"..."oh yeah"...and so on.
(1+ax)^x-1=(x/1!)*(ax)^1+((x)(x-1)/2!)*(ax)^2+… Lowest exponent possible x^2 is only given by first term, so neglect other powers (a=2,3) So given limit is a1/a2=2/3 Took 10 seconds
The way I would approach it is to work out (1+2t)^t = exp(ln(1+2t)*t). For small values of t: ln(1+2t)*t = t*(2t + O(t^2)) = 2*t^2 + O(t^3). This implies (1+2t)^t = 1 + 2 t^2 + O(t^3). Using a similar argument: we have that (1+3t)^t = 1 + 3 t^2 + O(t^3) for small values of t. Using the above approximations: we see that the expression we want to take limits for can be written as ((1+2t)^t-1)/((1+3t)^t-1) = (2 t^2 + O(t^3))/(3 t^2 + O(t^3)) = (2 + O(t))/(3 + O(t)) for small values of t. This expression tends to 2/3 as t tends to 0.
Wow that's kinda a tricky one, it looks terrifying and it appeared to be quite tough at the end🎉. But interesting. I just would recommend to everyone for future , if you see the alike expressions and you have to do some complex stuff with them, like differentiation, simply substitute that number which is different with some letter constant, say a, and at the end you can calculate what it would be if a=2 and a=3, just to be on the safe side. I'm also lazy, but if it comes to passing some exam , that error could cost too much, so I would definitely do that. Luckily that didn't effected the result😊
At 6:16 could we use binomial expansion? Since t goes to zero you can neglect "high" power of t (like t cubed) and, more importantly, the undetermined form vanish
Can't we use the following definition if exponential: Lim(n->inf)= (1+1/n)^(n)=e then if we have the limit (1+a/n)^n=(1+1/(n/a))^(n/a×a)=((1+1/(n/a))^(n/a))^a=e^a in the limit Then by forcing out X as common factor from the parenthesis we get ( e^2-1)/(e^3-1)
Girls : Taylor Swift & Billie Eillish are the best Boys : Lil Nas X & Jason Derulo are the best Mathematicians : PRIME NEWTONS' INTRO IS THE BEST!!!!!!!!!!!!!!!!!
@@PrimeNewtons oops, a typo, you know, thinking about one thing and typing something different Lol. Anyways I don't know that limit, why it's 1? Is it some known limit? Give me a hint, thanks
when you got 6+6t, shouldn't it be 6+9t? the answer would still be the same, but I thought that would be the right step
Why 9t? Understood why it was 6t
Doubling 3 will give 6, so why would it be 9?
Isso mesmo. Eu estava indo comentar sobre essa parte.
@@dutchie265 não é o dobro. quando se iguala as bases fica 3(1+3t)+3= 3+9t+3=6+9t. ele se confundiu porque por coincidência 2*2= 2². mas a generalização de d/dt [ ln(1+at) + at/(1+at)] = (2a+a²t)/(1+at)². ele pensou em (2a+2at)/(1+at)² erroneamente.
Yes its 6+9t i notice abkut it also from (3(1+3t)+3)/(1+3t)²
This is one of those "oh yeah!" Problems. I was never assigned anything like this in my calculus days. Nor have I assigned anything like this. So I looked at this problem and didnt even know where to start. But every step of the way, my response was "oh yeah"..."oh yeah"...and so on.
I have to do multiple variable analysis in university, but this videos really help in dealing with hard problems, thanks for the video!
That problem was completely crazy. It seemed to be a battle between you and the limit the whole time. As always, in the end, you won!
(1+ax)^x-1=(x/1!)*(ax)^1+((x)(x-1)/2!)*(ax)^2+…
Lowest exponent possible x^2 is only given by first term, so neglect other powers (a=2,3)
So given limit is a1/a2=2/3
Took 10 seconds
The way I would approach it is to work out (1+2t)^t = exp(ln(1+2t)*t). For small values of t:
ln(1+2t)*t = t*(2t + O(t^2))
= 2*t^2 + O(t^3). This implies
(1+2t)^t = 1 + 2 t^2 + O(t^3). Using a similar argument: we have that
(1+3t)^t = 1 + 3 t^2 + O(t^3) for small values of t.
Using the above approximations: we see that the expression we want to take limits for can be written as
((1+2t)^t-1)/((1+3t)^t-1)
= (2 t^2 + O(t^3))/(3 t^2 + O(t^3))
= (2 + O(t))/(3 + O(t)) for small values of t. This expression tends to 2/3 as t tends to 0.
This is like magic!!! Thank you.
For the last bottom it should be 6+9t from (3(1+3t)+3)/(1+3t)²
The answer is just the same tho cus it will just 0 at the end 😂😂
Correct
You are great! Regards from Poland!
Wow that's kinda a tricky one, it looks terrifying and it appeared to be quite tough at the end🎉. But interesting.
I just would recommend to everyone for future , if you see the alike expressions and you have to do some complex stuff with them, like differentiation, simply substitute that number which is different with some letter constant, say a, and at the end you can calculate what it would be if a=2 and a=3, just to be on the safe side. I'm also lazy, but if it comes to passing some exam , that error could cost too much, so I would definitely do that. Luckily that didn't effected the result😊
Very nice limit👍
The best math channel 👍
Very good lecture Sir. Thanks 👍
For applying L'Hospital:a rule, why use substitution?
By directly differentiating, and substitution, (nr. Treated as u^t-1, we get 2/3.
At 6:16 could we use binomial expansion? Since t goes to zero you can neglect "high" power of t (like t cubed) and, more importantly, the undetermined form vanish
We can do it. But binomial would be unnecessary here
@@abhishankpaul of course you don't need the whole expansion, you just need the first term (like when you use the Taylor series' trick)
@@davidcroft95 I got you, but what I meant to say is that binomial expansion will work, but that would be an overkill in this question
Thank you. It is very well done
Can't we use the following definition if exponential:
Lim(n->inf)= (1+1/n)^(n)=e then if we have the limit (1+a/n)^n=(1+1/(n/a))^(n/a×a)=((1+1/(n/a))^(n/a))^a=e^a in the limit
Then by forcing out X as common factor from the parenthesis we get ( e^2-1)/(e^3-1)
This one was a banger 🔥
Beatifull lim!!!
Girls : Taylor Swift & Billie Eillish are the best
Boys : Lil Nas X & Jason Derulo are the best
Mathematicians : PRIME NEWTONS' INTRO IS THE BEST!!!!!!!!!!!!!!!!!
2t(1+2t)^(t-1)
I didn't got the idea about infinite root of infinity, from where can one derive it to be zero?
It's 1 not zero
@@PrimeNewtons oops, a typo, you know, thinking about one thing and typing something different Lol. Anyways I don't know that limit, why it's 1? Is it some known limit? Give me a hint, thanks
nice
You are gorgeous.
Giant 1
we found t, not x
asnwer=1+2/5 isit
ah ah asnwer=2/3 isit
aNSwer , anyways wrong 😅