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
@ounaogot7 ай бұрын
Why 9t? Understood why it was 6t
@dutchie2657 ай бұрын
Doubling 3 will give 6, so why would it be 9?
@TheLukeLsd7 ай бұрын
Isso mesmo. Eu estava indo comentar sobre essa parte.
@TheLukeLsd7 ай бұрын
@@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.
@muhammadridholatif88177 ай бұрын
Yes its 6+9t i notice abkut it also from (3(1+3t)+3)/(1+3t)²
@JourneyThroughMath7 ай бұрын
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.
@BetaLoversYT7 ай бұрын
I have to do multiple variable analysis in university, but this videos really help in dealing with hard problems, thanks for the video!
@nothingbutmathproofs71506 ай бұрын
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!
@lukaskamin7557 ай бұрын
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😊
@cameronspalding97927 ай бұрын
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.
@Viesto19807 ай бұрын
You are great! Regards from Poland!
@muhammadridholatif88177 ай бұрын
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 😂😂
@PrimeNewtons7 ай бұрын
Correct
@juv70267 ай бұрын
(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
@michaelbaum67967 ай бұрын
Very nice limit👍
@saarike5 ай бұрын
This is like magic!!! Thank you.
@LORDLDUQ7 ай бұрын
The best math channel 👍
@davidcroft957 ай бұрын
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
@abhishankpaul6 ай бұрын
We can do it. But binomial would be unnecessary here
@davidcroft956 ай бұрын
@@abhishankpaul of course you don't need the whole expansion, you just need the first term (like when you use the Taylor series' trick)
@abhishankpaul6 ай бұрын
@@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
@surendrakverma5556 ай бұрын
Very good lecture Sir. Thanks 👍
@Edsonrsmtm7 ай бұрын
Beatifull lim!!!
@mirzatayerejepbayev83677 ай бұрын
Thank you. It is very well done
@AshokKumar-ul6dg6 ай бұрын
For applying L'Hospital:a rule, why use substitution? By directly differentiating, and substitution, (nr. Treated as u^t-1, we get 2/3.
@highestintheroom-mn7lt7 ай бұрын
This one was a banger 🔥
@Moj947 ай бұрын
nice
@lukaskamin7557 ай бұрын
I didn't got the idea about infinite root of infinity, from where can one derive it to be zero?
@PrimeNewtons7 ай бұрын
It's 1 not zero
@lukaskamin7557 ай бұрын
@@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
@artdeheer61157 ай бұрын
2t(1+2t)^(t-1)
@mustaphaballaji46287 ай бұрын
You are gorgeous.
@Subham-Kun5 ай бұрын
Girls : Taylor Swift & Billie Eillish are the best Boys : Lil Nas X & Jason Derulo are the best Mathematicians : PRIME NEWTONS' INTRO IS THE BEST!!!!!!!!!!!!!!!!!